The document discusses graphs in polar coordinates. It begins by reviewing how polar coordinates (r, ΞΈ) represent a point P, where r is the distance from the origin and ΞΈ is the angle relative to the x-axis. It then discusses how to convert between rectangular (x, y) and polar (r, ΞΈ) coordinates. The document explains that polar equations relate r and ΞΈ, giving the example of r = ΞΈ which graphs as an Archimedean spiral. It also discusses the graphs of constant equations like r = c, which is a circle, and ΞΈ = c, which is a line. The document concludes by explaining how to graph other polar equations by plotting points.
The document discusses parametric equations, which describe the motion of a particle in a plane by giving its position (x, y) at time t as functions of t, known as parametric equations. Examples are provided of parametric equations defining circles, ellipses, and other curves. The parameter does not need to be time. Slopes of parametric curves can be found from the derivatives of the parametric equations with respect to t. Standard parameterizations of functions y=f(x) are also discussed.
The document describes polar coordinates. Polar coordinates represent the location of a point P in a plane using two numbers: r, the distance from P to the origin O, and ΞΈ, the angle between the positive x-axis and a line from O to P. ΞΈ is positive for counter-clockwise angles and negative for clockwise angles. The polar coordinate (r, ΞΈ) uniquely identifies the location of P. Polar coordinates (r, ΞΈ) can be converted to rectangular coordinates (x, y) using the relations x=rcos(ΞΈ) and y=rsin(ΞΈ).
This document provides information about polar coordinates including:
- Relations between Cartesian and polar coordinates
- Sketching graphs in polar coordinates such as circles, cardioids, and roses
- Finding intersections of curves, slopes of tangents, and areas bounded by polar curves
- Computing arc lengths and surfaces of revolution generated by polar curves
It discusses key concepts like symmetry properties and provides examples of computing specific values related to polar curves.
35 tangent and arc length in polar coordinatesmath266
Β
The document discusses parametric representations of polar curves. It begins by explaining that a rectangular curve given by y=f(x) can be parametrized as x=t and y=f(t). It then explains that a polar curve given by r=f(ΞΈ) can be parametrized as x=f(ΞΈ)cos(ΞΈ) and y=f(ΞΈ)sin(ΞΈ). Examples are given of parametrizing the Archimedean spiral r=ΞΈ and the cardioid r=1+sin(ΞΈ). Formulas are derived for calculating the slope of the tangent line to a polar curve at a given point in terms of r and ΞΈ. Examples are worked out for finding the slope
The document discusses polar coordinates and graphs. It begins by explaining how polar coordinates (r, ΞΈ) track the location of a point P in the plane, where r is the distance from the origin and ΞΈ is the angle from the x-axis. It then provides the conversions between rectangular (x, y) and polar coordinates. The document gives examples of basic polar graphs for constant equations like r = c, which describes a circle, and ΞΈ = c, which describes a line. It concludes by explaining how to graph other polar equations using a polar graph paper.
The document discusses polar coordinates and graphs. Polar coordinates (r, ΞΈ) can be used to specify the location of a point P by giving the distance r from the origin and the angle ΞΈ. Conversion formulas allow changing between polar (r, ΞΈ) and rectangular (x, y) coordinates. Polar equations relate r and ΞΈ, and common ones like r = c (a circle) and ΞΈ = c (a line) are examined. Graphing polar equations involves plotting the r and ΞΈ values specified by the equation.
The document describes polar coordinates, which specify the location of a point P in a plane using two numbers: r, the distance from P to the origin O, and ΞΈ, the angle between the positive x-axis and the line from O to P. ΞΈ is positive for counter-clockwise angles and negative for clockwise angles. Conversion formulas between polar (r, ΞΈ) and rectangular (x, y) coordinates are provided. An example problem illustrates plotting points from their polar coordinates and finding the corresponding rectangular coordinates.
The document discusses graphs in polar coordinates. It begins by reviewing how polar coordinates (r, ΞΈ) represent a point P, where r is the distance from the origin and ΞΈ is the angle relative to the x-axis. It then discusses how to convert between rectangular (x, y) and polar (r, ΞΈ) coordinates. The document explains that polar equations relate r and ΞΈ, giving the example of r = ΞΈ which graphs as an Archimedean spiral. It also discusses the graphs of constant equations like r = c, which is a circle, and ΞΈ = c, which is a line. The document concludes by explaining how to graph other polar equations by plotting points.
The document discusses parametric equations, which describe the motion of a particle in a plane by giving its position (x, y) at time t as functions of t, known as parametric equations. Examples are provided of parametric equations defining circles, ellipses, and other curves. The parameter does not need to be time. Slopes of parametric curves can be found from the derivatives of the parametric equations with respect to t. Standard parameterizations of functions y=f(x) are also discussed.
The document describes polar coordinates. Polar coordinates represent the location of a point P in a plane using two numbers: r, the distance from P to the origin O, and ΞΈ, the angle between the positive x-axis and a line from O to P. ΞΈ is positive for counter-clockwise angles and negative for clockwise angles. The polar coordinate (r, ΞΈ) uniquely identifies the location of P. Polar coordinates (r, ΞΈ) can be converted to rectangular coordinates (x, y) using the relations x=rcos(ΞΈ) and y=rsin(ΞΈ).
This document provides information about polar coordinates including:
- Relations between Cartesian and polar coordinates
- Sketching graphs in polar coordinates such as circles, cardioids, and roses
- Finding intersections of curves, slopes of tangents, and areas bounded by polar curves
- Computing arc lengths and surfaces of revolution generated by polar curves
It discusses key concepts like symmetry properties and provides examples of computing specific values related to polar curves.
35 tangent and arc length in polar coordinatesmath266
Β
The document discusses parametric representations of polar curves. It begins by explaining that a rectangular curve given by y=f(x) can be parametrized as x=t and y=f(t). It then explains that a polar curve given by r=f(ΞΈ) can be parametrized as x=f(ΞΈ)cos(ΞΈ) and y=f(ΞΈ)sin(ΞΈ). Examples are given of parametrizing the Archimedean spiral r=ΞΈ and the cardioid r=1+sin(ΞΈ). Formulas are derived for calculating the slope of the tangent line to a polar curve at a given point in terms of r and ΞΈ. Examples are worked out for finding the slope
The document discusses polar coordinates and graphs. It begins by explaining how polar coordinates (r, ΞΈ) track the location of a point P in the plane, where r is the distance from the origin and ΞΈ is the angle from the x-axis. It then provides the conversions between rectangular (x, y) and polar coordinates. The document gives examples of basic polar graphs for constant equations like r = c, which describes a circle, and ΞΈ = c, which describes a line. It concludes by explaining how to graph other polar equations using a polar graph paper.
The document discusses polar coordinates and graphs. Polar coordinates (r, ΞΈ) can be used to specify the location of a point P by giving the distance r from the origin and the angle ΞΈ. Conversion formulas allow changing between polar (r, ΞΈ) and rectangular (x, y) coordinates. Polar equations relate r and ΞΈ, and common ones like r = c (a circle) and ΞΈ = c (a line) are examined. Graphing polar equations involves plotting the r and ΞΈ values specified by the equation.
The document describes polar coordinates, which specify the location of a point P in a plane using two numbers: r, the distance from P to the origin O, and ΞΈ, the angle between the positive x-axis and the line from O to P. ΞΈ is positive for counter-clockwise angles and negative for clockwise angles. Conversion formulas between polar (r, ΞΈ) and rectangular (x, y) coordinates are provided. An example problem illustrates plotting points from their polar coordinates and finding the corresponding rectangular coordinates.
A vector is a quantity with both magnitude and direction. There are two main operations on vectors: vector addition and scalar multiplication. Vector addition involves placing the tail of one vector at the head of another and drawing the third side of the resulting triangle or parallelogram. Scalar multiplication scales the length of a vector without changing its direction. Vectors can be represented using Cartesian components, where the magnitude and direction of a vector are given by its x, y, and z values relative to a set of perpendicular axes.
This document discusses converting between rectangular and polar coordinates and equations. It provides examples of converting specific points from rectangular to polar coordinates and vice versa, as well as converting equations from rectangular form to polar form and from polar form to rectangular form. Students are assigned problems from their textbook to practice these conversions.
Introduction To Polar Coordinates And Graphseekeeney
Β
Polar coordinates use r and ΞΈ instead of x and y, where r represents the radius or distance from a point to the pole, and ΞΈ represents the angle between the radius and the polar axis. To convert between rectangular and polar coordinates, r is calculated as the distance from the origin using the Pythagorean theorem, while ΞΈ is the angle measured counterclockwise from the polar axis. Polar graphs represent r as a function of ΞΈ, with ΞΈ as the independent variable instead of x as in rectangular graphs.
The document describes polar coordinates, which specify the location of a point P in a plane using two numbers: r, the distance from P to the origin O, and ΞΈ, the angle between the x-axis and a line from O to P measured counterclockwise. Conversion formulas between polar (r, ΞΈ) and rectangular (x, y) coordinates are provided. An example problem converts several polar coordinates to rectangular form and plots the points on a graph.
The document discusses polar equations and their use in describing curves. Polar equations are defined as equations involving the variables r and ΞΈ. Common polar equations like r = c, ΞΈ = c, r = Β±c*cos(ΞΈ), and r = Β±c*sin(ΞΈ) are presented along with examples of how they describe geometric shapes like circles and lines. In particular, the equations r = Β±c*cos(ΞΈ) and r = Β±c*sin(ΞΈ) always describe circles.
Polar coordinates provide an alternative way to specify the location of a point P in a plane using two numbers: r, the distance from P to the origin O, and ΞΈ, the angle between the x-axis and a line from O to P. ΞΈ is measured counter-clockwise from the x-axis and can be either positive or negative. A point P's polar coordinates (r, ΞΈ) uniquely identify its location. Polar and rectangular coordinates can be converted between each using the relationships x=r*cos(ΞΈ), y=r*sin(ΞΈ), and r=β(x2+y2).
The document discusses polar coordinates and graphs. It begins by explaining how polar coordinates (r, ΞΈ) track the location of a point P in the plane, where r is the distance from the origin and ΞΈ is the angle from the positive x-axis. It then provides the conversions between rectangular (x, y) and polar coordinates. The document introduces polar equations, showing how they relate r and ΞΈ, and gives examples of the constant equations r = c and ΞΈ = c, which describe a circle and line, respectively. It concludes by explaining how to graph other polar equations by plotting points using a polar graph paper.
This document discusses trigonometric functions and their properties. It defines periodic functions and their periods. It describes the amplitude of sine and cosine functions as half the difference between the maximum and minimum values. It discusses how transformations of a, b, h, and k values can stretch, compress, reflect, translate and shift the graphs of sine and cosine functions and how these affect their amplitudes and periods. Examples are provided to demonstrate identifying amplitudes, periods, phase shifts from transformed trigonometric functions.
The document discusses calculating the area swept out by a polar function r=f(ΞΈ) between two angles ΞΈ=A and ΞΈ=B. It introduces the integral formula for finding this area, which is β«f(ΞΈ)2dΞΈ from A to B. It then provides examples of using this formula to calculate the areas of different polar curves, such as a circle, a cardioid, and a curve tracing a circle twice.
This document provides an overview of 8 lessons on polar coordinates taught in a Further Pure Mathematics II course. The lessons cover key concepts such as plotting curves in polar form, converting between Cartesian and polar coordinates, determining maximum and minimum values of polar curves, finding equations of tangents, and calculating areas using polar coordinates. Practice problems are provided after each lesson covering the material through exercises and past exam questions.
This document provides an overview of polar coordinates including:
- Polar coordinates use (r, Ξ) notation where r is the distance from the origin and Ξ is the angle from the polar axis.
- Polar coordinates can be converted to rectangular coordinates using the equations x = r cos Ξ and y = r sin Ξ.
- Rectangular coordinates can be converted to polar coordinates by using the Pythagorean theorem to find r and trigonometric functions to find Ξ.
- Examples are provided for converting between polar and rectangular coordinates.
The document provides information about curve tracing, including important definitions, methods of tracing curves, and examples. It defines key curve concepts like singular points, multiple points, points of inflection, and asymptotes. The method of tracing curves involves analyzing the equation for symmetry, points of intersection with axes, regions where the curve does not exist, and determining tangents and asymptotes. Four examples are provided to demonstrate how to apply this method to trace specific curves and identify their properties.
BCA_Semester-II-Discrete Mathematics_unit-iv Graph theoryRai University
Β
This document defines key concepts in graph theory, including:
- A graph is defined as a pair (V,E) where V is the set of vertices and E is the set of edges.
- Examples of graph terminology include vertices, edges, walks, paths, circuits, connectivity, and components.
- Different types of graphs are discussed such as simple graphs, complete graphs, subgraphs, and induced subgraphs.
Coordinate systems (and transformations) and vector calculus garghanish
Β
The document discusses various coordinate systems and vector calculus concepts. It defines Cartesian, cylindrical, and spherical coordinate systems. It describes how to write vectors and relationships between components in different coordinate systems. It also covers vector operations like gradient, divergence, curl, and Laplacian as well as line, surface, and volume integrals. Examples are provided to illustrate calculating the gradient of scalar fields defined in different coordinate systems.
The document describes polar coordinates. Polar coordinates represent the location of a point P in a plane using two numbers: r, the distance from P to the origin O, and ΞΈ, the angle between the positive x-axis and a line from O to P. ΞΈ is positive for counter-clockwise angles and negative for clockwise angles. The polar coordinate (r, ΞΈ) uniquely identifies the point P. Conversions between polar coordinates (r, ΞΈ) and rectangular coordinates (x, y) are given by the equations x=r*cos(ΞΈ), y=r*sin(ΞΈ), and r=β(x2+y2).
This document defines higher-order outer and inner radial derivatives of set-valued maps. It provides the definitions, discusses the motivation and differences from other higher-order derivatives, and notes that the radial derivative considers global rather than just local properties of sets and maps. The document also reviews various notions of efficiency in vector optimization and how they relate to the concept of Q-minimality.
Interpreting Multiple Regressionvia an Ellipse Inscribed in a Square Extensi...Toshiyuki Shimono
Β
The document discusses interpreting multiple regression results geometrically using ellipses. It presents 3 theorems:
1) The multiple correlation coefficient is the ratio of line segments from the origin to a point P inside the ellipse and its projection P' onto the ellipse.
2) Regression coefficients are values of linear scalar fields defined on the axes at point P.
3) Partial correlation coefficients are values of affine functions defined on the longest line segments through P parallel to the axes.
The document defines vectors and discusses their geometric and algebraic representations. Geometrically, a vector has a magnitude and direction represented by an arrow. Algebraically, a vector in a plane can be represented by its coordinates (a1, a2) and in 3D space by coordinates (a1, a2, a3). Vectors can be added by placing them head to tail, subtracted by reversing one and adding, and scaled by a scalar number. The dot product of two vectors A and B yields a scalar value that geometrically equals the magnitudes of A and B multiplied by the cosine of the angle between them.
The document discusses vector fields and line integrals. Some key points:
- A vector field associates a vector with each point in a region, such as a velocity vector field showing wind patterns.
- Line integrals generalize the idea of integration over an interval to integration over a curve. The line integral of a function f over a curve C is defined as the limit of Riemann sums that multiply f by the length of curve segments.
- Line integrals can be used to calculate properties like work done by a force field or circulation in fluid flow. Their value does not depend on the parametrization of the curve C.
The document discusses transformations in geometry. It defines a geometric transformation as a bijective mapping between two geometries that maps points to points and lines to lines. Reflections, rotations, and translations are provided as examples of geometric transformations in Euclidean plane geometry. It is shown that reflections, rotations, and translations are isometries that preserve distance, angle measure, and area. The composition of transformations is also discussed, and it is shown that the composition of isometries is again an isometry.
A vector is a quantity with both magnitude and direction. There are two main operations on vectors: vector addition and scalar multiplication. Vector addition involves placing the tail of one vector at the head of another and drawing the third side of the resulting triangle or parallelogram. Scalar multiplication scales the length of a vector without changing its direction. Vectors can be represented using Cartesian components, where the magnitude and direction of a vector are given by its x, y, and z values relative to a set of perpendicular axes.
This document discusses converting between rectangular and polar coordinates and equations. It provides examples of converting specific points from rectangular to polar coordinates and vice versa, as well as converting equations from rectangular form to polar form and from polar form to rectangular form. Students are assigned problems from their textbook to practice these conversions.
Introduction To Polar Coordinates And Graphseekeeney
Β
Polar coordinates use r and ΞΈ instead of x and y, where r represents the radius or distance from a point to the pole, and ΞΈ represents the angle between the radius and the polar axis. To convert between rectangular and polar coordinates, r is calculated as the distance from the origin using the Pythagorean theorem, while ΞΈ is the angle measured counterclockwise from the polar axis. Polar graphs represent r as a function of ΞΈ, with ΞΈ as the independent variable instead of x as in rectangular graphs.
The document describes polar coordinates, which specify the location of a point P in a plane using two numbers: r, the distance from P to the origin O, and ΞΈ, the angle between the x-axis and a line from O to P measured counterclockwise. Conversion formulas between polar (r, ΞΈ) and rectangular (x, y) coordinates are provided. An example problem converts several polar coordinates to rectangular form and plots the points on a graph.
The document discusses polar equations and their use in describing curves. Polar equations are defined as equations involving the variables r and ΞΈ. Common polar equations like r = c, ΞΈ = c, r = Β±c*cos(ΞΈ), and r = Β±c*sin(ΞΈ) are presented along with examples of how they describe geometric shapes like circles and lines. In particular, the equations r = Β±c*cos(ΞΈ) and r = Β±c*sin(ΞΈ) always describe circles.
Polar coordinates provide an alternative way to specify the location of a point P in a plane using two numbers: r, the distance from P to the origin O, and ΞΈ, the angle between the x-axis and a line from O to P. ΞΈ is measured counter-clockwise from the x-axis and can be either positive or negative. A point P's polar coordinates (r, ΞΈ) uniquely identify its location. Polar and rectangular coordinates can be converted between each using the relationships x=r*cos(ΞΈ), y=r*sin(ΞΈ), and r=β(x2+y2).
The document discusses polar coordinates and graphs. It begins by explaining how polar coordinates (r, ΞΈ) track the location of a point P in the plane, where r is the distance from the origin and ΞΈ is the angle from the positive x-axis. It then provides the conversions between rectangular (x, y) and polar coordinates. The document introduces polar equations, showing how they relate r and ΞΈ, and gives examples of the constant equations r = c and ΞΈ = c, which describe a circle and line, respectively. It concludes by explaining how to graph other polar equations by plotting points using a polar graph paper.
This document discusses trigonometric functions and their properties. It defines periodic functions and their periods. It describes the amplitude of sine and cosine functions as half the difference between the maximum and minimum values. It discusses how transformations of a, b, h, and k values can stretch, compress, reflect, translate and shift the graphs of sine and cosine functions and how these affect their amplitudes and periods. Examples are provided to demonstrate identifying amplitudes, periods, phase shifts from transformed trigonometric functions.
The document discusses calculating the area swept out by a polar function r=f(ΞΈ) between two angles ΞΈ=A and ΞΈ=B. It introduces the integral formula for finding this area, which is β«f(ΞΈ)2dΞΈ from A to B. It then provides examples of using this formula to calculate the areas of different polar curves, such as a circle, a cardioid, and a curve tracing a circle twice.
This document provides an overview of 8 lessons on polar coordinates taught in a Further Pure Mathematics II course. The lessons cover key concepts such as plotting curves in polar form, converting between Cartesian and polar coordinates, determining maximum and minimum values of polar curves, finding equations of tangents, and calculating areas using polar coordinates. Practice problems are provided after each lesson covering the material through exercises and past exam questions.
This document provides an overview of polar coordinates including:
- Polar coordinates use (r, Ξ) notation where r is the distance from the origin and Ξ is the angle from the polar axis.
- Polar coordinates can be converted to rectangular coordinates using the equations x = r cos Ξ and y = r sin Ξ.
- Rectangular coordinates can be converted to polar coordinates by using the Pythagorean theorem to find r and trigonometric functions to find Ξ.
- Examples are provided for converting between polar and rectangular coordinates.
The document provides information about curve tracing, including important definitions, methods of tracing curves, and examples. It defines key curve concepts like singular points, multiple points, points of inflection, and asymptotes. The method of tracing curves involves analyzing the equation for symmetry, points of intersection with axes, regions where the curve does not exist, and determining tangents and asymptotes. Four examples are provided to demonstrate how to apply this method to trace specific curves and identify their properties.
BCA_Semester-II-Discrete Mathematics_unit-iv Graph theoryRai University
Β
This document defines key concepts in graph theory, including:
- A graph is defined as a pair (V,E) where V is the set of vertices and E is the set of edges.
- Examples of graph terminology include vertices, edges, walks, paths, circuits, connectivity, and components.
- Different types of graphs are discussed such as simple graphs, complete graphs, subgraphs, and induced subgraphs.
Coordinate systems (and transformations) and vector calculus garghanish
Β
The document discusses various coordinate systems and vector calculus concepts. It defines Cartesian, cylindrical, and spherical coordinate systems. It describes how to write vectors and relationships between components in different coordinate systems. It also covers vector operations like gradient, divergence, curl, and Laplacian as well as line, surface, and volume integrals. Examples are provided to illustrate calculating the gradient of scalar fields defined in different coordinate systems.
The document describes polar coordinates. Polar coordinates represent the location of a point P in a plane using two numbers: r, the distance from P to the origin O, and ΞΈ, the angle between the positive x-axis and a line from O to P. ΞΈ is positive for counter-clockwise angles and negative for clockwise angles. The polar coordinate (r, ΞΈ) uniquely identifies the point P. Conversions between polar coordinates (r, ΞΈ) and rectangular coordinates (x, y) are given by the equations x=r*cos(ΞΈ), y=r*sin(ΞΈ), and r=β(x2+y2).
This document defines higher-order outer and inner radial derivatives of set-valued maps. It provides the definitions, discusses the motivation and differences from other higher-order derivatives, and notes that the radial derivative considers global rather than just local properties of sets and maps. The document also reviews various notions of efficiency in vector optimization and how they relate to the concept of Q-minimality.
Interpreting Multiple Regressionvia an Ellipse Inscribed in a Square Extensi...Toshiyuki Shimono
Β
The document discusses interpreting multiple regression results geometrically using ellipses. It presents 3 theorems:
1) The multiple correlation coefficient is the ratio of line segments from the origin to a point P inside the ellipse and its projection P' onto the ellipse.
2) Regression coefficients are values of linear scalar fields defined on the axes at point P.
3) Partial correlation coefficients are values of affine functions defined on the longest line segments through P parallel to the axes.
The document defines vectors and discusses their geometric and algebraic representations. Geometrically, a vector has a magnitude and direction represented by an arrow. Algebraically, a vector in a plane can be represented by its coordinates (a1, a2) and in 3D space by coordinates (a1, a2, a3). Vectors can be added by placing them head to tail, subtracted by reversing one and adding, and scaled by a scalar number. The dot product of two vectors A and B yields a scalar value that geometrically equals the magnitudes of A and B multiplied by the cosine of the angle between them.
The document discusses vector fields and line integrals. Some key points:
- A vector field associates a vector with each point in a region, such as a velocity vector field showing wind patterns.
- Line integrals generalize the idea of integration over an interval to integration over a curve. The line integral of a function f over a curve C is defined as the limit of Riemann sums that multiply f by the length of curve segments.
- Line integrals can be used to calculate properties like work done by a force field or circulation in fluid flow. Their value does not depend on the parametrization of the curve C.
The document discusses transformations in geometry. It defines a geometric transformation as a bijective mapping between two geometries that maps points to points and lines to lines. Reflections, rotations, and translations are provided as examples of geometric transformations in Euclidean plane geometry. It is shown that reflections, rotations, and translations are isometries that preserve distance, angle measure, and area. The composition of transformations is also discussed, and it is shown that the composition of isometries is again an isometry.
The document discusses transformations in geometry. It defines a geometric transformation as a bijective mapping between two geometries that maps points to points and lines to lines. Reflections, rotations, and translations are provided as examples of geometric transformations in Euclidean plane geometry. It is shown that reflections, rotations, and translations are isometries that preserve distance, angle measure, and area. The composition of transformations is also discussed, and it is shown that the composition of isometries is again an isometry.
This document provides notes on vector spaces, which are fundamental objects in linear algebra. It begins with examples of vector spaces such as R2, R3, C2, C3 and defines vector spaces more generally as sets that are closed under vector addition and scalar multiplication and satisfy other properties like the existence of additive identities. It then provides several examples of vector spaces including the set of all n-tuples over a field, the set of all mΓn matrices, the set of differentiable functions on an interval, and the set of polynomials with coefficients in a field.
This document discusses four ways to represent functions: verbally, numerically, visually through graphs, and algebraically through explicit formulas. It provides examples of each type of representation and discusses key properties of functions like domain, range, and the vertical line test. The document also covers piecewise-defined functions and functions with symmetry properties like even functions whose graphs are symmetric about the y-axis.
The document defines equivalence relation and provides two examples. It then proves some properties about equivalence relations on real numbers. It proves mathematical induction for a formula relating sums and cubes. It proves properties about spanning trees and connectivity in graphs. It also proves that congruence modulo m is an equivalence relation by showing it satisfies the properties of reflexivity, symmetry, and transitivity. Finally, it explains the concepts of transition graphs and transition tables for representing finite state automata.
This document discusses vector fields and gradient fields. It provides examples of different types of vector fields including:
- Wind velocity vectors showing patterns in San Francisco Bay
- Ocean currents and airfoil flow as examples of velocity vector fields
- The gravitational force field as an example of a force field
- Component functions that can be used to express vector fields as the sum of scalar functions
- Gradient vector fields defined as the gradient of a scalar function of variables
Contour maps are shown to have gradient vectors perpendicular to level curves, with longer vectors where curves are closer together.
LΓmite y continuidad de una funciΓ³nΒ en el Espacio R3
Derivadas parciales
Diferencial total.
Gradientes
Divergencia yΒ Β Rotor
Plano tangente y recta normal
Regla de la cadena
Jacobiano.
Extremos relativos
Multiplicadores de Lagrange
Integral en lΓnea
Teorema de Gauss
Teorema de Ampere
Teorema de Stoke
Teorema de Green
Polyhedral and algebraic method in computational geometrySpringer
Β
This chapter introduces geometric fundamentals that will serve as the basis for topics covered later. It lays the foundations of projective geometry, which allows for a simpler formulation of statements compared to affine geometry. Projective spaces are defined by extending affine spaces so that parallel lines intersect at a point at infinity. Linear transformations between vector spaces induce projective transformations between the corresponding projective spaces. Any two maximal flags in a projective space can be mapped to each other by some projective transformation.
This document discusses probabilistic diameter and its properties. It defines probabilistic diameter (DA) as a distribution function that represents the probability that the distance between any two points in a set A is less than some value t. It presents several properties of probabilistic diameter including: (1) DA is a distribution function; (2) DA = H if A contains a single point; and (3) if A is a subset of B, then DA β₯ DB. It also defines probabilistic distance between two sets A and B as another distribution function (FAB) and establishes some of its properties.
Chapter 4: Vector Spaces - Part 1/Slides By PearsonChaimae Baroudi
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This document defines vectors and vector spaces. It begins by defining vectors in 2D and 3D space as matrices and describes operations like addition, scalar multiplication, and subtraction. It then defines a vector space as a set of vectors that satisfies 10 axioms related to these operations. Examples of vector spaces include the set of 2D and 3D vectors, sets of matrices, and sets of polynomials. The document also defines subspaces and proves that the span of a set of vectors in a vector space forms a subspace.
Section 18.3-19.1.Today we will discuss finite-dimensional.docxkenjordan97598
Β
Section 18.3-19.1.
Today we will discuss finite-dimensional associative algebras and their representations.
Definition 1. Let A be a finite-dimensional associative algebra over a field F . An element
a β A is nilpotent if an = 0 for some positive integer n. An algebra is said to be nilpotent if
all of its elements are.
Exercise 2. Every subalgebra and factor algebra of a nilpotent algebra are nilpotent. Con-
versely, if I β A is a nilpotent ideal, and the quotient algebra A/I is nilpotent, then so is
A.
Example. The algebra of all strictly lower- (or upper-) triangular nΓn matrices is nilpotent.
Proposition 3. If the algebra A is nilpotent, then An = 0 for some n β Z+, that is the
product of any n elements if the algebra A equals 0.
Proof. Let B β A be the maximal subspace for which there exists n β Z+ such that Bn = 0.
Note that B is closed under multiplication, i.e. B β Bk for any k β Z+. Assume B 6= A
and choose an element a β A\B. Since aBn = 0, there exists k β Z+ such that aBk 6β B.
Replacing a with a non-zero element in aBk we obtain aB β B. Recall that there exists
m β Z+ so that am = 0. Now, let us set C = B βγaγ. Then we have
Cmn = 0
which contradicts the definition of the subspace B. οΏ½
Unless A is commutative, the set of all nilpotent elements of A does not have to be an
ideal (in general, not even a subspace). On the other hand, if I,J are nilpotent ideals in A,
then so is
I + J = {x + y |x β I,y β J} .
Therefore, there exists the maximal nilpotent ideal which contains every other nilpotent ideal
of A.
Definition 4. The radical of A is the maximal nilpotent ideal of A it is denoted rad(A).
The algebra A is semisimple if rad(A) = 0.
If char(F) = 0, there exists an alternative description of semisimple algebras. Consider
the regular representation of the algebra A
Ο: A β L(A), Ο(a)(b) = ab,
and define a βscalar productβ on A via
(a,b) = tr(Ο(ab)) = tr(Ο(a)Ο(b)).
One can see that (Β·, Β·) is a symmetric bilinear function on A satisfying (ab,c) = (a,bc).
Definition 5. For any ideal I β A, its orthogonal complement Iβ₯ is defined as
Iβ₯ = {a β A |(a,i) = 0 for all i β I} .
Proposition 6. For any ideal I β A, its orthogonal complement Iβ₯ β A is also an ideal.
Proof. For any a β A, x β Iβ₯, and y β I we have
(xa,y) = (x,ay) = 0 and (ax,y) = (y,ax) = (ya,x) = 0.
οΏ½
1
2
Proposition 7. If A is an algebra over a field F with char(F) = 0, then every element a β A
orthogonal to all of its powers is nilpotent.
Proof. Let a β A be such that
(a,an) = tr Ο(a)n+1 = 0 for all n β Z+.
Let L β F be the splitting field of the characteristic polynomial f(x) of the operator Ο(a).
Then, over L we have
f(x) = tk0
sβ
i=1
(tβΞ»i)ki,
where Ξ»i are distinct for i = 1, . . . ,s, and
tr Ο(a)n+1 =
sβ
i=1
kiΞ»
n+1
i = 0.
Letting n run through the set 1, . . . ,s, the above equation yields a system of s homogeneous
linear equations in variables k1, . . . ,ks. The determinant of this system equals
Ξ»21 . . .Ξ»
2
nV (Ξ»1, . . . ,Ξ»n),
where V (Ξ»1.
Section 18.3-19.1.Today we will discuss finite-dimensional.docxrtodd280
Β
Section 18.3-19.1.
Today we will discuss finite-dimensional associative algebras and their representations.
Definition 1. Let A be a finite-dimensional associative algebra over a field F . An element
a β A is nilpotent if an = 0 for some positive integer n. An algebra is said to be nilpotent if
all of its elements are.
Exercise 2. Every subalgebra and factor algebra of a nilpotent algebra are nilpotent. Con-
versely, if I β A is a nilpotent ideal, and the quotient algebra A/I is nilpotent, then so is
A.
Example. The algebra of all strictly lower- (or upper-) triangular nΓn matrices is nilpotent.
Proposition 3. If the algebra A is nilpotent, then An = 0 for some n β Z+, that is the
product of any n elements if the algebra A equals 0.
Proof. Let B β A be the maximal subspace for which there exists n β Z+ such that Bn = 0.
Note that B is closed under multiplication, i.e. B β Bk for any k β Z+. Assume B 6= A
and choose an element a β A\B. Since aBn = 0, there exists k β Z+ such that aBk 6β B.
Replacing a with a non-zero element in aBk we obtain aB β B. Recall that there exists
m β Z+ so that am = 0. Now, let us set C = B βγaγ. Then we have
Cmn = 0
which contradicts the definition of the subspace B. οΏ½
Unless A is commutative, the set of all nilpotent elements of A does not have to be an
ideal (in general, not even a subspace). On the other hand, if I,J are nilpotent ideals in A,
then so is
I + J = {x + y |x β I,y β J} .
Therefore, there exists the maximal nilpotent ideal which contains every other nilpotent ideal
of A.
Definition 4. The radical of A is the maximal nilpotent ideal of A it is denoted rad(A).
The algebra A is semisimple if rad(A) = 0.
If char(F) = 0, there exists an alternative description of semisimple algebras. Consider
the regular representation of the algebra A
Ο: A β L(A), Ο(a)(b) = ab,
and define a βscalar productβ on A via
(a,b) = tr(Ο(ab)) = tr(Ο(a)Ο(b)).
One can see that (Β·, Β·) is a symmetric bilinear function on A satisfying (ab,c) = (a,bc).
Definition 5. For any ideal I β A, its orthogonal complement Iβ₯ is defined as
Iβ₯ = {a β A |(a,i) = 0 for all i β I} .
Proposition 6. For any ideal I β A, its orthogonal complement Iβ₯ β A is also an ideal.
Proof. For any a β A, x β Iβ₯, and y β I we have
(xa,y) = (x,ay) = 0 and (ax,y) = (y,ax) = (ya,x) = 0.
οΏ½
1
2
Proposition 7. If A is an algebra over a field F with char(F) = 0, then every element a β A
orthogonal to all of its powers is nilpotent.
Proof. Let a β A be such that
(a,an) = tr Ο(a)n+1 = 0 for all n β Z+.
Let L β F be the splitting field of the characteristic polynomial f(x) of the operator Ο(a).
Then, over L we have
f(x) = tk0
sβ
i=1
(tβΞ»i)ki,
where Ξ»i are distinct for i = 1, . . . ,s, and
tr Ο(a)n+1 =
sβ
i=1
kiΞ»
n+1
i = 0.
Letting n run through the set 1, . . . ,s, the above equation yields a system of s homogeneous
linear equations in variables k1, . . . ,ks. The determinant of this system equals
Ξ»21 . . .Ξ»
2
nV (Ξ»1, . . . ,Ξ»n),
where V (Ξ»1.
1) Derivatives relate the rates of change of position, velocity, and acceleration. Velocity is the derivative of position and measures rate of change of displacement. Acceleration is the derivative of velocity and measures the rate of change of velocity.
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This document discusses differentials and how they relate to differentiable functions. Some key points:
1. The differential of an independent variable x is defined as dx, which is equal to the increment Ξx. The differential of a dependent variable y is defined as dy = f'(x) dx, where f'(x) is the derivative of the function.
2. Differentials allow approximations of changes in a function using derivatives, such as estimating errors or finding approximate roots.
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This document discusses various methods of proving mathematical propositions, including direct proof, indirect proof, proof by contradiction, proof by cases, and proof by mathematical induction. It provides examples to illustrate each method. Direct proof involves directly deducing the conclusion from the given statements, while indirect proof establishes an equivalent proposition. Proof by contradiction assumes the statement is false and arrives at a contradiction. Proof by cases examines all possible cases of the hypothesis. Mathematical induction proves a statement for all natural numbers based on an initial case and assuming the statement holds for k implies it holds for k+1.
This document discusses various methods of proving mathematical propositions, including direct proof, indirect proof, proof by contradiction, proof by cases, and proof by mathematical induction. It provides examples to illustrate each method. Direct proof involves directly deducing the conclusion from the given statements, while indirect proof establishes an equivalent proposition. Proof by contradiction assumes the negation of the statement to be proved and arrives at a contradiction. Proof by cases considers all possible cases of the hypothesis. Mathematical induction proves a statement for all natural numbers based on proving it for the base case and assuming it is true for some arbitrary case k.
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This document discusses three types of vectors: numeric vectors, geometric/physical vectors, and functions. Numeric vectors are lists of numbers. Geometric/physical vectors have magnitude and direction, like directed line segments representing displacements. Functions can also be viewed as vectors. All three types of vectors can be added, subtracted, and multiplied by numbers. Numeric vectors correspond to geometric vectors through their components in a coordinate system. Forces are represented as geometric vectors with magnitude and direction.
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2. ii
QUOTES
The Prophet Muhammad (β«ο·Ίβ¬ β peace be upon him) said: βAcquire
knowledge and impart it to the people.β β (Sunan Tirmidhi, Hadith
107)
βWhere there is matter, there is geometryβ
Johannes Kepler
βMeaning is important in mathematics and geometry is an important
source of that meaningβ.
David Hilbert
3. iii
Preface
The writer aims to thank to the Almighty God, Allah, because of His bless
and grace, this module titled βTransformation Geometryβ can be finished.
Moreover, the writer expresses gratitude to the Mathematics Department
lecturers, peculiarly, geometry field lecturers.
This module is intended as a completion of geometry literatures which uses
English as the medium of instruction. Therefore, it can be used a reference
for International Class Program students. Related to the contents, it
consists of several transformations on the Euclidean plane, i.e. reflection,
halfturn, translation, and rotation. Besides that, there are some concepts
related to the transformations concerned in this module namely bijective
function, isometry, and the composition of two transformations.
The writer hopes this module is certainly useful for everyone, particulary
for Mathematics Department students. However, critiques and advices are
emphatically needed for the refinement of this module in future.
4. iv
LIST OF CONTENTS
Page Title ...................................................................................i
Quotes ...................................................................................ii
Preface .................................................................................. iii
List of contents ................................................................................. iv
Transformation ...................................................................................1
Reflection ...................................................................................8
Isometry ...................................................................................15
Composition of Transformations ..................................................................................22
Halfturn ..................................................................................29
Translation ..................................................................................37
Rotation ..................................................................................48
References ..................................................................................56
5. 1
CHAPTER I
TRANSFORMATION
The discussion of transformation geometry commences with the introduction of the concept
of function that have been studied in the subject of calculus. The concept underlies the topic of
transformation geometry, for example, one of the postulates in Euclidean geometry, i.e. every
angle on a plane is associated with exactly one real number.
The function which is discussed here is restricted to the function having domain and origin
in the form of V (V is the Euclidean plane). The function definition is given as follows:
In general, a function is notated with the letter π. If π is a function from V to V that associates
every x β V to y β V, it can be written as π¦ = π( π₯) where π₯ is called the pre-image of π¦ by π
and π¦ is called the image of π₯ by π. The origin and the range of the function are π.
In the calculus course, the types of functions are described, however, here only three types
are discussed, namely:
a. Surjective Function
A function f is called a surjective function if for every y element V, there exists x element
V such that π( π₯) = π¦. And to show that a function is surjective, we must show that every
element (image) has pre-image. In other words, for every element in codomain has pair in
domain V.
b. Injective Function
A function π is called injective if for every a and b element of the domain, where a β b
then f(a) β f(b). The statement is equivalent to βif f(a) = f(b) then a = bβ. To show that a
function is injective, we must show that for every pair a and b element domain, if a β b then
we show f(a) β f(b), or if f(a) = f(b) then we show a = b.
Definition:
A function of V to V is a mapping that associates every element of V to exactly one element
of V
.
6. 2
c. Bijective Function
A function f is called bijective if f is surjective function and injective function. So, to show
that a function is bijective, we must show that the function is both surjective function as well
as injective function.
In Mathematics subject in Junior High School (SMP) and Senior High School (SMA) we
have learned about symmetry, rotation, translation, and dilatation. All we have learned are the
equivalent of bijective and those are transformations that will be discussed.
Whereas, transformation geometry term can be interpreted as a branch of geometry that
discusses transformation, but it can also be interpreted as a geometry which is based on the
transformation.
The following discussion presents the geometry in the first interpretation, but at the same time it
leads to the second interpretation.
To show that a mapping of V to V is a transformation, then the steps which should be
consecutively undertaken are checking whether:
1. The mapping is a function.
2. The mapping is surjective.
3. The mapping is injective.
Surjective means that if β B β V, β A β V such that T(A) = B.
B = mapping from A by T, and
A = pre-image of B by T.
T(A) = Aβ²
Injective means that if A and B are elements of the domain, then β(A β B) βΉ (Aβ²
β Bβ²
)β,
where T(A) = Aβ and T(B) = Bβ.
The statement is equivalent to β(Aβ² = Bβ²) βΉ (A = B)β
So, to show that a function is injective, then it must be shown that for each pair of elements of
the domain A and B, if A β B then we must show that Aβ²
β Bβ²
, or if Aβ² = Bβ² then it must be
shown that A = B.
Definition
A transformation on plane V is a bijective function of which both the domain and the
codomain are V. Where V is the Euclidean plane
7. 3
A surjective and injective function is a bijective function. Therefore to show whether the
function is bijective, it must be shown that whether the function is both surjective and injective.
Example 1:
Suppose A β V. There is a mapping T of which both the domain and the codomain are V.
T: V β V is defined as follows:
1) T(A) = A.
2) If P β A, then T(P) = Pβ² with Pβ² the middle point π΄πΜ Μ Μ Μ .
Show that the mapping T is a transformation.
Solution:
The image of point A is the point A itself.
Take a point R β A on V.
Since V is the Euclidean plane, then there is one line passes through the points A and R.
It means that there is only one line segment AR, so that there is exactly one point S with S β π΄π Μ Μ Μ Μ
, such that AS = SR.
Since R is an arbitrary point, it means that for each X β V, there is Y β V, where Y = T(X).
Therefore T is a function
Is T surjective?
In other words, does every point in V has a pre-image?
To answer these questions, it must be shown that for arbitrary point Y β V, is there X β V so that
T(X) = Y.
8. 4
According to the first condition, if Y = A, its pre-image is A itself, because T(A) = A. If Y β A,
since V is the Euclideanplane, then there is exactly one X with X β π΄πΜ Μ Μ Μ such that AY = YX. It
means that X is pre-image of the point Y.
Thus, every point in V has a pre-image which implies that T is a surjective mapping.
Is T injective?
To show that T is injective, take any point P, Q β V with P β Q.
- The first case if P, Q and A are not collinear.
It will be shown that the position of Pβ²
= T(P) and Qβ²
= T(Q)
Suppose Pβ²
= Qβ².
Since Pβ² β π΄πΜ Μ Μ Μ and Qβ²
β π΄πΜ Μ Μ Μ , then π΄πΜ Μ Μ Μ and π΄πΜ Μ Μ Μ has two intersection points, namely point A
and point Pβ or Qβ. It means π΄πΜ Μ Μ Μ and π΄πΜ Μ Μ Μ coincide Since Q β π΄πΜ Μ Μ Μ and π΄πΜ Μ Μ Μ = π΄πΜ Μ Μ Μ then Q β π΄πΜ Μ Μ Μ ,
resulting in point P, Q and A are collinear.
It is contrary to the fact that P, Q and A are not collinear. It means that the assumption that
Pβ²
= Qβ² is not true, and it should be Pβ²
β Qβ².
So if P β Q then Pβ²
β Qβ², which means injective.
- The second case is if P, Q and A are collinear (it is also injective)
From the proof above, it turns out that T is an injective mapping.
9. 5
Because T is a surjective and injective mapping then T is bijective. So T is a transformation.
Example 2:
Given relation T [(x, y)] = (2x + 1, y β x).
Show that this mapping is a transformation.
Solution:
If P(x, y) then Pβ²
= T(P) = (2x + 1, y β x). It means that the domain of T is the whole plane V.
Is T surjective?
Take an arbitrary point π΄(x, y), is π΄ a domain of T ?
Suppose that B(xβ, yβ) is a domain of point A, then certainly T(B) = A or T[(xβ, yβ)] = (x, y) β
(2xβ² + 1, yβ² β xβ²) = (x, y)
we get π₯β²
=
π₯β1
2
πππ π¦β²
=
2π¦+π₯β1
2
.
So B = (
xβ1
2
,
2yΒ±1
2
), hence T((
xβ1
2
,
2yΒ±1
2
) = (x, y).
Since (xβ, yβ) always exists for each (x, y) then B (domain of A) always exists meaning that
T(B) = A. Because A is an arbitrary point in V, then each point in V has domain which means
that T is surjective.
Is T injective?
Take points P(x1, y1) and Q(x2, y2) where P β Q
Is Pβ β Qβ ?
Suppose that Pβ = Qβ then ( 2x1 + 1, y1 β x1) = (2x2 + 1, y2 β x2)
Since π₯1 = π₯2 and π¦1 = π¦2 then P = Q. It is contrary to the fact that P β Q.
It means that the assumption that Pβ = Qβ is false and it should be Pβ β Qβ.
It is now proven that if P β Q then Pβ β Qβ. So, T is injective.
Therefore T is injective and surjective, in other words, T is a transformation
12. 8
CHAPTER II
REFLECTION
In senior high school physics, we have learned about the properties of reflection. It says
that if an object is x units in the front of a mirror, then the image of the object is also located as
far as x units behind the mirror, and if the object is located on the mirror, its image will coincide
with the object. It is further discussed geometrically in this reflection section:
Reflection in line s is notated Ms. Line s is called the axis of reflection or mirror line.
As shown in the previous section, to show whether reflection is a transformation, it must
be shown whether the reflection is a function that is surjective and injective.
To show that a reflection is a transformation, the steps which should be undertaken are
answering the following questions:
1. Is reflection a function?
Based on its definition, reflection is a function from V to V.
2. Is reflection surjective?
Take arbitrary point A 'on V. If Aββ s. Geometrically, A is the element of V so s become
the axis π΄π΄β²Μ Μ Μ Μ Μ (since V is an Euclidean plane).
It means that the MS(A) = Aβ implying that every A' has pre-image. Then, M is surjective
3. Is reflection injective?
Take two arbitrary points A, B β V where A β B.
There are three possibilities, namely:
a. A β s dan B β s
It means Ms(A) = Aβ = A and Ms(B) = Bβ = B
Since A β B it means Aββ Bβ
Then M is injective.
b. A β s and B β s
It means ππ (π΄) = π΄β = π΄ and Ms(B) = Bβ such that s is the axis π΅π΅β²Μ Μ Μ Μ Μ .
Because A β s and Bβ β s , so that Aββ Bβ
Then M is injective.
c. A β s and B β s
Assume that Ms(A) = Ms(B) or Aβ = Bβ. Since π΄π΄β²Μ Μ Μ Μ Μ β₯ s and π΅π΅β²Μ Μ Μ Μ Μ β₯ s , so π΅π΄β²Μ Μ Μ Μ Μ β₯ s , thus it
is obtained that from point Aβ, two distinct lines can be created perpendicular to line s
which is impossible. Then the assumption that MS(A) = Ms(B) or A '= B' is false. Thus,
it should be Aββ Bβ.
Thus, if A β B then Aββ Bβ.
Definition:
Reflection in line π is a function ππ that is defined for each point P on plane V
as follows:
i. If P β s so Ms = P.
ii. If P β s so Ms (P) = Pβ , in a way such that s is the axis of ππβ²Μ Μ Μ Μ Μ .
13. 9
Then, M is injective.
Since MS is a function that is both surjective and injective, Ms is transformation.
Suppose that s: ax + by + c = 0 , and P (x,y), where Ms(P) = Pβ(xβ, yβ).
β P(x,y) s
βPβ(xβ,yβ)
If P β s so ππβ²Μ Μ Μ Μ Μ β₯ s , then:
π¦β²βπ¦
π₯β²βπ₯
=
π
π
.................... (i)
If Q is the midpoint of ππβ²Μ Μ Μ Μ Μ , then Q(
π₯+π₯β²
2
,
π¦+π¦β²
2
) lies on the line s.
So a(
π₯+π₯β²
2
) + b(
π¦+π¦β²
2
) + c = 0.................(ii)
From the equation (i) and (ii), it is obtained that:
xβ = x -
2π(ππ₯+ππ¦+π)
π2+ π2
yβ = y -
2π(ππ₯+ππ¦+π)
π2+ π2
so, if s: ax + by + c = 0, and P(x,y), then Ms(P) = Pβ(xβ,yβ) where
xβ = x -
2π(ππ₯+ππ¦+π)
π2+ π2
yβ = y -
2π(ππ₯+ππ¦+π)
π2+ π2
Theorem:
Every reflection in a line is a transformation
14. 10
Example 1:
If on V there is an orthogonal axis system with A (1,3) and B (-2,1). Determine the equation of
the line s so Ms (A) = B!.
Solution:
Ms(A) = B means that s is the axis π΄π΅Μ Μ Μ Μ . If T is the midpoint of π΄π΅Μ Μ Μ Μ , then s passes through the point
T and perpendicular to π΄π΅Μ Μ Μ Μ . mπ΄π΅β‘ββββ =
2
3
=> mS = -
3
2
, and T (-
1
2
, 2).
So, the line s : y β 2 = -
3
2
(x +
1
2
) or
s : 3x + 2y - 2
1
2
= 0 or s : 6x + 4y β 5 = 0
Example 2 :
Suppose line s : 2x β 3y + 5 = 0
a. Determine Ms(A) if A(2,-5)
b. Determine Ms(O)
Solution:
Given line s : 2x β 3y + 5 = 0 => a=2 , b=-3 , and c=5
a. A(2,-5) , Ms(A) = Aβ(xβ,yβ)
xβ = x -
2π( ππ₯+ππ¦+π)
π2+ π2
xβ = 2 -
2.2[2.2+(β3).(β5)+5]
22+ (β3)2
= 2 -
96
13
=
β70
13
yβ = y -
2π( ππ₯+ππ¦+π)
π2+ π2
yβ = -5 -
2.(β3)[2.2+(β3).(β5)+5]
22+ (β3)2
= -5 -
β144
13
=
79
13
Thus, Ms(A) = Aβ (
β70
13
,
79
13
)
c. Ms(O) = Oβ(xβ,yβ)
xβ = x -
2π( ππ₯+ππ¦+π)
π2+ π2
15. 11
= 0 -
2.2[2.0+(β3).0+5]
22+ (β3)2
= 0 -
20
13
=
β20
13
yβ = y -
2π( ππ₯+ππ¦+π)
π2+ π2
= 0 -
2.(β3)[2.0+(β3).0+5]
22+ (β3)2
= 0 -
(β30)
13
=
30
13
Thus, Ms(O) = Oβ (
β20
13
,
30
13
)
Exercises :
1. Suppose line g = {(x,y)|y = x}
a. If A(2,-3), determine Mg(A).
b. If Bβ(-3,5), determine pre-image of Bβ by Mg.
c. If P(x,y) any point, determine Mg(P).
2. Suppose h = {(x,y)|y = 2}
a. If C(3,β2 ), determine Cβ.
b. If Dβ(2,-4), determine pre-image D, by Mh.
c. If P(x,y), determine Pβ.
3. Suppose s = {(x,y)|x = -3}
a. If A(4,1), determine Aβ = Ms(A).
b. Determine the coordinate of point C if Ms(C) = (-2,7).
c. If P(x,y) is any point, determine Ms(P).
4. Suppose line l = {(x,y)|2x + 3y = 11}
a. Determine Ml(O).
b. Determine Ml(E) with E(1,2).
c. If F(x, 2x-1), determine the coordinate F if Ml(F) = F.
5. Suppose line s = {(x,y)|2x + y = 1} and t = {(x,y)|x = -2}. Find the equation of line sβ = Mt(s).
6. Suppose line t, circle l with center D, and β ABC as shown below :
16. 12
l
B t
A
a. Draw Mt(β ABC)
b. Draw Mt(l)
7. If lines g = {(x,y)|y = 1}, h ={(x,y)|y = x} and k ={(x,y)|x = 3}. Find the equation of the
following lines :
a. Mg(h) c. Mh(g)
b. Mg(k) d. Mh(k)
8. Mk is a reflection that connects point A(4,8) to point B(8,0). Determine the equation (image)
of circle (x + 1)2
+ (y β 3)2
= 9, if it is reflected to the line k.
9. If two points P and Q. Draw a line t so that Mt(P) = Q and determine Mt(Q).
10. There are an orthogonal axis system in V, point A(1,3), and point B(-2,-1). Determine the
equation of a line of g so that Mg(A) = B.
11. Given two parallel lines g and h, points A and B as shown in the figure. Draw the shortest
path from A to B providing that it must be reflected on g then on h.
g
h
12. If g = {(x,y)|y = -x} and h ={(x,y)|3y = x + 3}
Show that whether point A(-2,-4) lies on line hβ = Mg (h).
13. If line g ={(x,y)|6x β 3y + 1=0} and a point A(k,2).
Find the value of k if Mg(A) = A.
14. Two walls form an angle as shown in the figure of which it is formed by line k and line l. A
ball is located in the point A. Sketch where the ball should be directed such that if it is reflected
on k and on l, it will be bounced back to A.
β D
C
Aβ
β B
17. 13
k
15. Given line g = {(x,y)|3x β y + 4 = 0} and h = {(x,y)|2x + 3y = 6}.
Determine the equation of line gβ = Mh(g), and hβ = Mg(h).
Aβ
l
18. 14
CHAPTER III
ISOMETRY
In daily life, many events or movements are transformations such as the movement of a
table and the opening or closing of a door. The movement of a table from one place to another
and the opening or closing of door do not alter the length and the width of table or doors except
the position of the table or the door. Such kind of transformation is called isometry.
Definition
A transformation π is an isometry if for every pair of points π, π satisfies πβ²
πβ²
=
ππ, where πβ²
= π(π) and πβ²
= π(π).
If π΄β²
= ππ (π΄), π΅β = ππ (π΅), then π΄βπ΅β = π΄π΅. (prove!)
Theorem
Any reflection on a line is an isometry.
As being previously discussed, the result of a reflection is preserving the lenght of
segment or the distance between two points, thus reflection is isometry.
Besides preserving the distance between two points, isometry also has the properties
as follows:
a. Mapping aline into a line
Suppose g is a line and π is an isometry, it will be proved that πβ² = π(π) is a line.
Put the points π΄ and π΅ on the line π (π΄ β π and π΅ β π).
Suppose π(π΄) = π΄β² and π(π΅) = π΅β², create a line h through point π΄β²and π΅β². It will be
proved that the line β = πβ²
Take an arbitrary point π on π such that it forms π΄ππ΅, and let πβ² = π(π). On the line π,
π΄π + ππ΅ = π΄π΅.
Because π is an isometry then π΄β²π΅β² = π΄π΅, π΄β²πβ² = π΄π, πβ²π΅β² = ππ΅.
Suppose πβ lies on outside of β, then at βπ΄β²πβ²π΅β² must be satisfied that π΄β²πβ² + πβ²π΅β² > π΄β²π΅β²
since π΄β²π΅ β² = π΄π΅, π΄β²πβ² = π΄π, π΄π + ππ΅ > π΄π΅
19. 15
It contradicts to the assumption that π΄π + ππ΅ = π΄π΅.
It means that the assumption that πβ² lies on outside of the h is not true. Suppose πβ² lies
on β or π΄β²πβ²π΅β².
So πβ β β
The reverse direction is proved in the same way by assuming that Q' is
any point on h.
Since π is a transformation, it means that it satisfies the surjective properties, then there
is a π such that π(π) = πβ². Suppose π lies outside of π. Using the triangle inequality, it
can be proved that π should be on g, so that πβ²
= π(π) must be on πβ²
= π (π). Thus, it
means that ββ² β πβ².
Since πβ² β β and β β πβ² then β = πβ²
b. Preserving the size of the angle between two lines
Take the three points π΄, π΅, and πΆ which are not collinear.
π΄β² = π(π΄), π΅β² = π(π΅) and πΆβ² = π (πΆ)
See the π₯π΄π΅πΆ. According to (a), since the AB and BC are straight lines so π΄β²π΅β² and
π΅β²πΆβ² are also straight lines.
Because π is an isometry then π΄β²π΅β² = π΄π΅, π΅β²πΆβ² = π΅πΆ, and π΄β²πΆβ² = π΄πΆ.
It means that π₯π΄π΅πΆ β π₯π΄β²π΅β²πΆ β²(s.s.s), which also means that the vertices of the
triangles are in the same position and the same magnitude.
Thus, isometry preserves the angles.
20. 16
c. Preserving the parallels of two lines
Given Line π // β, π β² = π (π) and ββ² = π (β).
It will be proved that π β²// ββ²
Suppose the line π 'intersect ββ² at the point πβ, then πβ² β πβ² and πβ² β ββ². Since π is a
transformation then for πβ² β ββ² there must exist π so that π(π) = πβ² where π β π and
π β β. It then implies π and β intersect at point π.
It contradicts to the assumption that π // β implying the assumption that π 'intersects ββ²
is wrong, it should be πβ²// ββ². One consequence of that property is that if πββ then
πβ²ββ β², where T is an isometry, πβ² = π(π), and ββ² = π(β). If π₯π΄π΅πΆ are reflected to the
line g, the map is π₯π΄β²π΅β²πΆβ², or ππ(π₯π΄π΅πΆ) = π₯π΄β²π΅β²πΆβ²
A reflection in the line π maps the π₯π΄π΅πΆ on π₯π΄β²π΅β²πΆβ². If the π₯π΄π΅πΆ with the order of the
π΄ β π΅ β πΆ is opposite to the clockwise direction, then the image which is π₯π΄β²π΅β²πΆβ having
the order π΄β² β π΅β² β πΆβ² is clockwise direction.
21. 17
Definition
i. A transformation π maintains an orientation if for any three points which
are not collinear (π, π, π ) have the same orientation as the orientation of
the triple points (πβ², πβ², π β²)
ii. A transformation T reverses an orientation if the orientation of any three
points which are not collinear (π, π, π ) is not the same as the orientation
of (πβ²
, πβ²
, π β²) where πβ² = π(π), πβ²
= π(π), and π β² = π(π ).
Meanwhile, the rotation about the center of rotation π maps π₯πππ on π₯πβ²πβ²π β². If on the
π₯πππ , the direction of π β π β π is clockwise direction then its image π₯πβ²πβ²π β², has also
clockwise direction πβ² β πβ² β π β².
Concerning the further discussion of isometry phenomenon above, we shall
introduce the concept of orientation of the pair of three points which are not collinear.
Suppose (π΄, π΅, πΆ) is the pair three points which are not collinear. Then through π΄,
π΅, and πΆ, there is exactly one circle πΌ. We can circumnavigate πΌ started from π΄, then in
π΅, then in πΆ and ended up back at π΄.
If the circumferential direction is the same as the clockwise direction, it is said
that the pair of three points (π΄, π΅, πΆ) has opposite orientations to clockwise direction or
negative orientation.
It means that the reflection of the line g that maps π₯π΄π΅πΆ into the π₯π΄β²π΅β²πΆβ²,
(π΄, π΅, πΆ) has negative orientation, and (π΄β², π΅ β², πΆβ²) has positive orientation. While in the
rotation with the center of rotation π which maps π₯πππ into the π₯πβ²πβ²π β², (π, π, π ) has
a positive orientation and (πβ², π β², π β²)βs orientation is also positive.
Therefore
Moreover, we can classify isometry into direct isometry and indirect isometry (opponent
isometry) by looking at the following definition:
22. 18
Definition
A transformation refers to direct isometry if the transformation preserves the
orientation, and it is called opponent isometry if the transformation reverses the
orientation.
An isometry refers to direct isometry if the isometry preserves orientation and
called opponent isometry if the isometry changes the orientation. Thus, since the
reflection changes the rotation, it means that the reflection is opponent isometry, while
the rotation is a direct isometry since it maintains orientation.
Furthermore, exactly one property holds, i.e. any isometry is either direct
isometry only or an opponent isometry only (not both).
Example 1 :
π is a transformation defined by π (π) = (π₯ β 7, π¦ + 4) for all points π (π₯, π¦) β π.
Show whether π is an isometry?
Solution:
Suppose that πβ = π(π) = (π₯ β 7, π¦ + 4) β π(π₯, π¦) β π
Suppose also that Point π΄(π₯1, π¦1) and (π₯2, π¦2) with π΄ β π΅.
It means that π΄β = π(π΄) = ( π₯1 β 7, π¦1 + 4) and
π΅β = π(π΅) = (π₯2 β 7, π¦2 + 4)
π΄π΅ = β( π₯2 β x1)2 + ( π¦2 β π¦1)2
π΄β²π΅β² = β{( π₯2 β 7) β (x1 β 7)}2 + {( π¦2 + 4) β ( π¦1 + 4)}2
= β( π₯2 β x1)2 + ( π¦2 β π¦1)2
Since π΄βπ΅β = π΄π΅, so π isometry
Example 2
Suppose π is a transformation that is defined for all points π(π₯, π¦) as π(π) = (β π¦, π₯).
a. Is π an isometry?
b. If π isometry, is it direct isometry or opponent isometry? Answer:
Solution:
a. Given πβ = π(π) = (π¦, βπ₯) β π(π₯, π¦) β π
Suppose that point π΄(π₯1, π¦1) and (π₯2, π¦2) with π΄ β π΅.
π΄π΅ = β( π₯2 β x1)2 + ( π¦2 β π¦1)2
π΄β²π΅β² = β( π¦2 β π¦1)2 + {(βπ₯2) β (βx1)}2
= β( π¦2 β π¦1)2 + (βπ₯2 + x1)2
= β( π₯2 β x1)2 + ( π¦2 β π¦1)2
evidently π΄βπ΅β = π΄π΅
23. 19
Because π΄βπ΅β = π΄π΅, so π isometry.
b. To check whether π is a direct isometry or opponent isometry, take any three points
which are not collinear, for example π(0.0), π΄(1,2) and π΅(4.1).
By using transformation π(π) = (π¦, βπ₯) ) β π(π₯, π¦) β π then it is obtained that πβ =
(0,0), π΄β = (β2,1), and π΅β = (β1,4).
Since the orientation (π, π΅, πΆ) is positive, and the orientation (πβ², π΄β², π΅β²) is also
positive, then π is a direct isometry.
Exercises
1. π is a transformation defined by π(π) = (3π₯ + 5,2 β 4π¦) for all points π(π₯, π¦) β
π.
Show that T is an isometry.
2. Suppose the points π΄(1, β1), π΅(4,0), πΆ(β4,1) and π·(β2, π). If an isometry π with
π(π΄) = πΆ and π(π΅) = π·, find the value of π.
3. Prove that the transformation with the formula:
[
π₯β²
π¦β²
] = [
3/5 β4/5
β4/5 β3/5
] [
π₯
π¦] is an isometry.
4. Suppose π₯π΄π΅πΆ by isometry π is mapped into π₯π΄β²π΅β²πΆβ², prove that π₯π΄π΅πΆ β
π₯π΄β²π΅β²πΆβ².
5. Given a line π. π a function defined for each point of π in the plane π as follows:
i. If π β π then π(π) = π
ii. If π β π then π(π) = πβ such that πβ² is the midpoint of the line segment of
orthogonal from π to π.
a. Is π a transformation?
b. Is π an isometry?
c. If there are two points π΄ and π΅ so that π΄β²π΅ β² = π΄π΅ where π΄β² = π (π΄),
π΅ β² = π (π΅), what can be interpreted about π΄ and π΅?
6. Suppose the π line and the points π΄, π΄β², π΅ and πΆ with π΄β² = ππ(π΄). By only using a
ruler without ascale, sketch point π΅β = ππ(π΅) and πΆβ = ππ(πΆ).
24. 20
7. Given lines π , π‘, π’; points π΄ and π΅ as in the following figure. π is an isometry
where π΅ = π(π΄) and π’ = π(π ). If π‘βπ , painting π‘β² = π(π‘)
8. Suppose a line π and and a circle πΌ. Prove that ππ(πΌ) = πΌβ with πΌβ is also a circle.
9. Suppose lines π, πβ, β, ββ and π where πβ = ππ(π) and ββ = ππ(β). If πβ//ββ prove
that π//β.
10. Suppose lines π, β, and ββ where ββ = ππ(β). Verifying the truth of the following
expressions:
a. If ββ//β, then β//π
b. If ββ = β, then β = π.
c. If βββ β = {π΄}, then π΄ β π.
11. Suppose line g and two points π΄ and π΅ as shown in the following figure:
a. by using an appropriate isometry, determine a point π β π so that π΄π + ππ΅ as
short as possible.
b. If π β π is different to the point π, prove that π΄π + ππ΅ > π΄π + ππ΅.
12. Suppose a circle πΌ = {(π₯, π¦) | (π₯ β 2) 2 + (π¦ β 3) 2 = 4}. π is an isometry
mapping point π΄(2,3) on π΄β²(1,7).
Find the equation of the set π(πΌ).
Is the map of I also circle? Why?
25. 21
13. In the following figure, there are three points which are not collinear, it is π, π, π . π
and π are isometry where πβ² = π(π), πβ² = π(π), π β² = π(π ), and π" = π(π), πβ² =
π(π), π β² = π(π ).
What kind of isometry π and π are those?
14. Isometry π maps point π΄ to π, point π΅ to π and πΆ to π . If π is an opponent isometry,
find (sketching) the position of point π .
15. Find the coordinates of the point π on the π₯ axis, measure of β π΄ππ = β π΅ππ, if
π΄ = (0,3) and π΅ = (6,5)
16. Suppose line g and points π΄, π΅ and π΄β where π΄β = ππ(π΄), and π΄π΅β‘ββββ //π. By using
only one ruler, determine the coordinate of point π΅β² = ππ(π΅).
26. 22
CHAPTER IV
COMPOSITION OF TRANSFORMATIONS
A. Composition Two Transformations
If F and G are respectively a transformation, then the composition of the two transformations is
defined as follows:
What about the composition of F and G, is it a transformation? To answer the question, whether
the composition of two transformations is also a transformation, then the following steps must
be solved consecutively:
1. Checking whether the composition of two transformations is functions
2. Checking whether the composition of two transformations is surjective
3. Checking whether the composition of two transformations is injective
Therefore, suppose the composition of F and G is H, or H = G. F.
β’ If F and G are functions, it is clear that H is also a function.
β’ Is H surjective?
Take arbitrary YβV. Is there XβV such that H (X) = Y? Since G is transformation, for every
YβV there is ZβV such that G (Z) = Y. Similarly, since F is transformation then for every ZβV,
there is XβV that F (X) = Z.
From πΊ(π) = π, it is obtained that πΊ[πΉ(π)] = π or (πΊ β πΉ)(π) = π, so π»( π) = π meaning
that π» is surjective.
β’ Is π» injective?
Take arbitrary π, π elements of π where π β π
Suppose ( π) = π»( π) , then πΊ[ πΉ( π)] = πΊ[ πΉ( π)]
Since πΊ is injective then πΉ(π) = πΉ(π), and since πΉ injective then π = π. It is contrary to which
it is known that π β π. It means that the assumption that π»(π) = π»(π) is not true. Therefore,
it should be π»(π) β π»(π) implying π» is injective.
Since π» is surjective as well as injective, then π» is bijective. Thus, π» is a transformation.
Definition
Suppose F and G are two transformations where F: V β V and G: V β V, then the
composition of F and G are notated G.F defined as (G.F) (P) = G [F (P)], β P βV.
27. 23
Theorem
The composition of two transformations is a transformation.
Example 1
Suppose that transformation π1[(π₯, π¦)] = (π₯ + 2, π¦) and π2[(π₯, 2π¦). If π is the composition of
π1 and π2, find π.
Solution:
π is the transformation π1 and π2 then :
π[(π₯, π¦)] = (π1 β π2 )(π₯, π¦)
= π2[π1 (π₯, π¦)]
= π2 [(π₯ + 2, βπ¦)]
= (π₯ + 2, β2π¦)
So the transformation π is [(π₯, π¦)] = (π₯ + 2, β2π¦) .
The Properties of The Composition of The Transformation
It has been discussed previously that the composition of two transformations is a
transformation. It means that the composition of transformations is closed. Furthermore, the
composition of transformations is also associative, but not commutative.
B. Inverse Transformation
If Mg(P) = Pβ²
, then Mg. Mg(P) = P or Mg
2
(P) = P.
So, M2
is a transformation that describes each point onto itself. This transformation is called the
identity transformation is symbolized by the letter I.
So that, I(P) = P, βP.
If T is a transformation, then T. I(P) = T[I(P)] = T(P), βP.
So, T. I = I and I. T(P) = I[T(P)] = T(P), βP.
Thus I. T = T. Implied T. I = I. T = T
Theorem
If T1, T2, and T3 are the transformation then T1[T2.T3]=[T1.T2].T3
28. 24
So that,the identity transformation I is the number 1 in the set of transformations with
multiplication operation among these transformations. In the set of real numbers, by
multiplication operations, each transformation T has inverse Q such that T.Q = I = Q.T.
If there is an invers transformation of T, then the transformation of T is written T-1
,so that T.T-
1
=T-1
.T = I
Proof:
Suppose π is a transformation. We define the equivalent πΉ as follows:
Suppose π is an element of π, where π is an euclidean field. Since π is a transformation, then π
is bijective implying that Aβ π, such that π(π΄) = π.
We define then πΉ(π) = π΄. meaning that πΉ(π) is the pre-image of π, such that from π(π΄) = π,
π[πΉ(π)] = π or (π. πΉ)(π) = πΌ(π), for every π element of π. So πΉπ = πΌ, so that ππΉ = πΉπ = πΌ.
Now it shall be proved that L is a transformation. From this definition, it is clear that F is
surjective. Suppose πΉ(π1) = πΉ(π2) and suppose π(π΄1) = π1, π(π΄2) = π2 where πΉ(π1) = π΄1
and πΉ(π2) = π΄2
Since T is a transformation, then π΄1 = π΄2, and it is also obtained that π1 = π2 such that F is
injective. Thus, it proves that πΉ is injective.
So πΉ is a transformation.
The transformation πΉ is called the inverse transformation of π and it is denoted by πΉ = πβ1
.
Proof:
Suppose T is a transformation with two inverses, namely S1 and S2 then
( π β π1 )( π) = ( π1 β π)( π) = πΌ( π) for all π and
Theorem
Each transformation π has an inverse
Theorem
Transformation has exactly one inverse.
29. 25
( π β π2 )( π) = ( π2 β π)( π) = πΌ( π) for all π
so, ( π β π1 )( π) = ( π β π2 )( π) β π [ π1 ( π)] = π [π2 ( π)]
because π is injective , then π1 ( π) = π2 ( π) for all π
So, π1 = π2 = π
Proof:
A reflection in the line π is Mg
For X π π then Mg (X) = X, So Mg . Mg (X) = Mg (X) = X = I(X)
So, Mg . Mg = I
Thus , the Mgβ1
= ππ
For X β π , Mg (X) = Xβ, so the π axis ππΜ Μ Μ Μ , then Mg . Mg (X) = Mg (Xβ) = X = I(X)
with π axis ππΜ Μ Μ Μ . So, Mg . Mg = I atau Mg
-1
= Mg
Proof:
Since (T.S) -1
is the inverse of (T.S) then (T.S) -1
.(T.S) = I. Meanwhile (S-1
.T-1
) (T.S) = S-1
.T-
1
.T.S = S-1
.I.S = S-1
.S = I
Since every transformation has only one inverse, then (TOS) -1
= S-1
T-1
Therefore the inverse of the composition transformations is the composition of the inverses of
each transformations in reverse order.
Example 1
In an orthogonal axis XOY system, transformations F and G are defined as follows
For β P (x, y), F (P) = (x + 2.1 / 2 y) and G (P) = (x-2,2y). (FG) (P) = F [G (P)] = F [(x-2,2y)] =
(x, y) = P
Theorem
Inverse of any reflection in a line is a reflection itself.
Definition
A transformation which has inverse which is the transformation itself is called an
involution.
Theorem
If π and π are transformations, then ( π β π) = πβ1
β πβ1
30. 26
While (GF) (P) = G [F (P)] = G [(x + 2.1 / 2 y) = (x, y) = P. So (FG) (P) = (GF) (P ) = P = I (P),
β P or FG = GF = I
Thus F and G are the mutual transformations of each other denoted by G = F-1
or F = G-1
Example 2
On an orthogonal axis system, the line g = {f (x, y) | y = x} and h = {(x, y) y = 0}
Find P such that (Mh.Mg) (P) = R with R (2,7)
Solution:
Let P (x, y)
(Mh.Mg) (P) = R => Mh (Mh.Mg) (P) = Mh (R)
β (Mh Mh Mg) (P) = Mh(R)
β (Mh Mh ) (Mg) (P) = Mh(R)
β (Mg) (P) = Mh(R)
β Mg.Mg (P) = Mg.Mh(R)
β P = Mg.Mh(R)
implying that P (x,y) = Mg.Mh(2,7) = Mg(2,-7) = (-7,2).
So the coordinate of the point P is (-7,2)
Exercises
1. Given lines g and h. A point K is the intersection of g and h, as well as the points P and Q points
on g and h. Sketch:
β’P
g
h β’Q
K
31. 27
a. A = Mg [Mh(P)]
b. B = Mh [Mg (P)]
c. C = Mh [Mh (P)]
d. D = Mg [Mh (K)]
e. R so Mh [Mg (R)] = Q
f. Do Mg.Mh = Mh.Mg ? Why?
2. Suppose that T and S are isometries, check whether the following statements below are true and
give the reason.
a. TS is an isometry
b. TS = ST
c. If g is a line, then gβ = (TS) (g) is also a line.
d. If g//h, and gβ = (TS) (g), hβ=(TS) (h), then gβ//hβ
3. Given two intersecting lines g and h, skecth
a. K such that Mg [Mb (k)] = g
b. b m such that Mh [Mg (m)] = g
c. N such that Mh [Mg (n)] line divides the acute angle between g and h.
4. The line g is the x-axis of an orthogonal axis system and h {(x, y | y = x}. Define:
a. The line equation Mh [Mg (g)]
b. P '' = Mh [Mg (P)], with P (0,3)
c. Qββ= Mg [Mb (Q)], with Q (3,-1)
d. Rββ= Mg. Mh (R) with R (x,y)
e. The magnitude of <RQR "when O is the origin
5. Let g is x-axis, and h = {(x, y | y = x}. S is a mapping defined as follows. If Pβg then S (P)
= P, And if Pβg then S (P) is the midpoint of the perpendicular line from P to G.
a. Prove that S is a transformation.
b. If P (x, y) any point, determine the coordinate of point S.Mg (P)
c. Check whether S. Mg = Mg.S.
d. Check whether S. Mh = Mh.S.
6. If g = (x, y) | y = 0} and h {(x, y) | y = x} and S is a transformation defined as question 5,
whereas A (2, -8) and P (x, y) , Determine the coordinates of the following points
a. Mg Mh S(A).
b. Mg.S .Mh (A).
32. 28
c. S Mg.S. (A).
d. Mh.S .Mg (P).
e. S2
.Mh (P).
f. S Mg
2
(P)
7. Suppose that g and h are two lines which are perpendicular to each other. A, B, and C are
three points so that Mg (A) = B and Mh (A) = C. Determine the following points.
a. Mg
3
(A)
b. Mh Mg Mh (A)
c. Mh Mg Mh Mh Mg (A)
d. Mg
2
Mh
3
(A)
8. Simplify.!
a. (WgVhMg)-1
b. (Mh Vh Wg Mg)-1
9. Suppose the transformations T1 [(x,y)] = (-x,y) and T2 [(x,y)]= (x,
1
2
y). Find the formula for
T2. T1 then if T 1= T2. T1, find the equation T (g) if g = {(x, y) | x + y = 0}.
What is T2. T1 = T1. T2?
10. If two different lines g and h intersect at point P, prove that Mg Mh (A) = P if and only if A
= P
11. It is known that g // h and points P, Q are neither on g nor on h.
a. Sketch Pββ = Mg Mh(P) and Qββ= Mg Mh(P)
b. What is the form of quadrilateral PPββQββQ ?
c. Prove your opinion!
33. 29
CHAPTER V
HALFTURN
Halfturn is a special case of rotation, where the rotation angle is 180Β°
. Since, halfturn
has special characteristic, it is discussed earlier. In the previous section, it has been discussed
that a reflection is an involution. Another example of an involution is a halfturn surrounding a
point. One halfturn reflects every point in a plane figure at a certain point.
Therefore, halfturn is called a reflection about point, and that point is the center of the
halfturn.
Definition
Halfturn about a point π΄ is a mapping ππ΄ that is defined for each point π on a plane as follows :
i. If π = π΄ , then ππ΄( π) = π.
ii. If π β π΄ , thenππ΄( π) = πβ², where π΄ as the center point of ππβ²Μ Μ Μ Μ Μ
Since halfturn is also a reflection of a point, and reflection is a transformation, then it can
be said that halfturn is a transformation.
Theorem
Halfturn is a transformation.
Suppose π΄(π, π),ππ΄ map point π(π₯, π¦) to πβ²(π₯β²
, π¦β²
), thenππ΄( π) = πβ² where π΄ is the center point
of ππβ²Μ Μ Μ Μ Μ so
π₯+π₯β²
2
= π and
π¦+π¦β²
2
= π.
It is obtained that π₯β²
= βπ₯ + 2π and π¦β²
= βπ¦ + 2π.
Thus, if π΄ = (π, π)and π (π₯, π¦), then ππ΄( π) = (2π β π₯, 2π β π¦).
Theorem
If π and β intersect and are perpendicular at point π΄, then ππ΄ = ππ πβ.
34. 30
Proof :
Since π β₯ β, we can make a system of orthogonal axis where π as axis π₯ and β as axis π¦, and π΄
is used as the point of origin. It must be proved that each point π (π₯, π¦), satisfies ππ΄(π) =
ππ πβ(π).
Let π (π₯, π¦) β π΄and ππ΄(π) = πβ²β²(π₯1, π¦1).
Since π΄ is the center point ππβ²β²Μ Μ Μ Μ Μ , so (0,0) = (
π₯1+π₯
2
,
π¦1+π¦
2
), so π₯1 = βπ₯ and π¦1 = βπ¦, therefore,
ππ΄( π) = (βπ₯, βπ¦).
While (ππ πβ)(π) = ππ[ πβ(π)] = ππ[(βπ₯, π¦)] = (βπ₯, βπ¦)
It turns out that ππ΄(π) = (ππ πβ)(π) = (π₯, π¦)
Thus, ππ΄ = ππ πβ
Theorem
If π and β are two lines perpendicular to each other, then ππ πβ = πβ ππ
Proof
If π = π΄ (Look at the figure of the previous theorem), so ππ πβ( π) = ππ( π) = π. Whereas
πβ ππ(π) = πβ( π) = π, so ππ πβ(π) = πβ ππ(π). if π β π΄, then ππ πβ = ππ΄, while
πβ ππ( π) = πβ[(π₯, π¦)] = (βπ₯, βπ¦) = ππ΄(π)
It turns out that ππ πβ = πβ ππ = ππ΄
So,ππ πβ = πβ ππ.
35. 31
Theorem
If ππ΄ is halfturn, then ππ΄
β1
= ππ΄.
Proof:
Suppose that π and β are two lines perpendicular to each other and intersect at point π΄,
then ππ πβ = ππ΄, implying that ππ΄
β1
= (ππ πβ)
β1
= πβ
β1
ππ
β1
= πβ ππ = ππ πβ = ππ΄. So
ππ΄
β1
= ππ΄.
Reflection in line π which is defined as ππ( π) = π when π β π, and ππ( π) = πβ²
. with
π is the axis ππΜ Μ Μ Μ when the π β π. When we look in general, so for every π β π the image of
point π is the point itself. Such point is called invariant point of the reflection.
Definition:
Point π΄ is called the invariant transformation of T, if it is satisfied that π( π΄) = π΄
It can be seen that reflection has many invariant points that are infinite, meanwhile
halfturn just has one invariant point, i.e. the center of the halfturn.
It has been discussed in the previous section that isometry is a transformation which maps
a line into a line. When a line by a transformation has image in the form of a line, such
transformation is called a collineation.
Based on the above understanding, any isometry is a collineation. Since halfturn is an
isometry, it is also a collineation. Among collineations, one of them is dilatation, defined as
follows :
One example of collineation which is dilatation is halfturn. The example was verified
by the following theorem :
Proof :
Definition:
A collineation (β) is called dilatation, if for every line π, it satisfies the property
βπ//π.
Theorem:
Suppose ππ΄ is a halfturn and π is a line, if π΄ β π, then ππ΄(π)//π
36. 32
Suppose π β π then π΄ is the midpoint of the segment ππβΜ Μ Μ Μ Μ with πβ²
= ππ΄(π).
Suppose π β π then π΄is the midpoint of the segment ππβ²Μ Μ Μ Μ Μ with πβ²
= ππ΄(π).
Since βπ΄ππ β βπ΄πβ²πβ² then πππβ²πβ² a parallelogram, implying that ππ // πβ²π". Thus, π // ππ΄(π).
Example 1 :
Suppose that two lines π and β are not parallel, π΄ is a point not located neither on π nor β.
Determine all points π on π and all points π on β such that π΄ is the midpoint of the segment ππΜ Μ Μ Μ .
Solution :
Take a point π β π. Sketching πβ²
= ππ΄(π). Then πβ²
= ππ΄(π) passes through πβ² where ππ΄ =
π΄πβ², πβ² // π. If πβ² intersecting β in π, then draw a line YA intersecting π in π. Then π and π are
the pair of the points which seems to be exactly the only one pair
a. Prove that π and π is the only pair that satisfy the condition.
b. If we didnβt use πβ²
= ππ΄(π) but we used ββ²β²
= ππ΄(β), could we get another pair?
Proof :
Suppose π is the center of a halfturn, and π is a line. We must prove that:
a. π π(π) // π
b. π π π π = πΌ, where πΌ is an identity transformation
Theorem :
A halfturn is a dilatation which has involutoric characteristic.
37. 33
Prove:
a. It means that π π( π) = πβ² is a line.
Suppose π΄ β π, π΅ β π, then π΄β²
β πβ²
, and ππ΄ = ππ΄β²
, ππ΅ = ππ΅β²
, while π (β π΄ππ΅) =
π(β π΄β²
πβ²
π΅β²
). Since βΏ ππ΄π΅ β βΏ ππ΄β²π΅β², π΄π΅π΄β²π΅β² is a parallelogram implying πβ²// π.
b. Since π π π π( π΄) = π π( π΄β²) = π΄ for all points π΄ β π, then π π π π( π) = πΌ( π).
So π π π π = πΌ meaning that π π is an involution.
The Composition of halfturns
The characteristics of the composition of halfturns are classified according to their centers
and whether there is an invariant point.
Proof:
Suppose π΄ and π΅ are the centers of the halfturns. Suppose π = π΄π΅, both β and π are
perpendicular line to π΄π΅ in π΄ and π΅ respectively, then:
ππ΄ π π΅ = (πβ ππ)(ππ ππ)
= [(πβ ππ)ππ]ππ
= [πβ (ππ ππ)]ππ
= πβ πΌ ππ
= πβ ππ
Suppose that π is the invariant point of SASB, then ππ΄ π π΅(π) = π implying (πβ ππ) (π) = π or
ππ(π) = πβ(π).
Suppose ππ(π) = π1
Theorem
The composition of two halfturns with different centers does not have invariant
point.
38. 34
If π β π1, then each β and π is ππ1 axis and since a line segment has exactly one axis, it must
be β = π. It is certainly not possible since π΄ β π΅.
If π = π1 then ππ(π) = π1 and πβ(π) = π1 , So π β β and π β π meaning that π and β
intersect at point π which is not possible because β // π.
Indeed, there can not be a point π such that π πΎ(π) = πβ(π) or ππ΄ π π΅ = π.
So ππ΄ π π΅ doesnβt have a fixed point.
Proof:
Suppose there are two halfturns π π· and π πΈ, so π π·(π΄) = π΅ and π πΈ (π΅) = π΄.
So, π π·( π΄) = π πΈ (π΅), then π π·[π π·(π΄)] = π π·[π πΈ(π΄)], then π΄ = π π· π πΈ(π΄). If π· and πΈ are
two different points, it means that π΄ is a fixed point π π· π πΈ. πΌ which is not possible. Therefore,
there is no more than one halfturns that maps π΄ to π΅. The only halfturn is π π(π΄) = π΅ where π is
the midpoint of π΄π΅Μ Μ Μ Μ .
Example:
Given πΈ = {( π₯, π¦)|π₯2
+ 4π¦2
= 16}, π΄(4, β3), and π΅(3,1).If π is the π axis, show that π΄ β
ππ π π΅(πΈ)?
Solution:
It is known that (π πS π΅)β1
= S π΅
β1
. M π
β1
= S π΅ ππ
If π(π₯, π¦), then ππ( π) = (π₯, βπ¦) then S π΅( π) = (2.3 β π₯, 2.1 β π¦) = (6 β π₯, 2 β π¦)
π΄ β ππ π π΅(πΈ) β π π΅ ππ(π΄) β πΈ
π π΅ ππ( π΄) = π π΅[ππ(4, β3)] = π π΅(4,3) = (2, β1)
Since (2, β1) β πΈ, then (ππ π π΅)
β1
( π΄) β ( πΈ)
In a similar way, we can define a set of maps if the equations are known.
In the latest example we know that π β ππ π π΅( πΈ) if and only if (ππ π π΅)
β1
( π) β ( πΈ).
If π(π₯, π¦) so (ππ π π΅)
β1
( π) = (6 β π₯, 2 + π¦), then (ππ π π΅)
β1
( π) β ( πΈ) if and only if
(6 β π₯, 2 + π¦) β {( π₯, π¦)|π₯2
+ 4π¦2
= 16}.
Theorem
If π΄ and π΅ are two different points, then there is only one and a halfturn that
maps π΄ to π΅.
39. 35
So it must be (6 β π₯)2
+ 4(2 + π¦)2
= 16. π( π₯, π¦) β ππ π π΅( πΈ) if and only if π( π₯, π¦) β
{( π₯, π¦)|π₯2
+ 4π¦2
β 12π₯ + 16π¦ + 36 = 0}
Therefore π₯2
+ 4π¦2
β 12π₯ + 16π¦ + 36 = 0 is the equation of the image of πΈ by
transformation of ππ π π΅.
Exercises
1. Suppose three distinct points A, B, P are not collinear, sketch
a. ππ΄(π)
b. ππ΄ π π΅( π)
c. R so π π΅( π ) = π
d. ππ΄
2
( π)
2. Given the line g and point A, π΄ β π
a. Draw lines πβ²
= ππ΄( π), why ππ΄( π) is a line?
b. Prove that πβ²
// π
3. Given βABC and a parallelogram WXYZ. There is a point K which lies outside the triangle
βABC and the parallelogram WXYZ.
a. Sketh πK(βABC)!
b. Find a point J so SJ(WXYZ) = WXYZ
4. If A = (2,3) determine!
a. S A (C) if C (2,3) b. S A (D) if D (-2,7)
c. S A
-1
(E) if E (4, -1) d. S A (P) if P (x, y)
5. If C = (-4.3) and g = {(x, y) | y = x}, determine!
a. M g S c (2.4) b. M g S c (P) if P = (x, y)
c. (M g S c)
-1
(P) d. Is M g S c = S c M g? Explain
6. Give the implication from the following expressions :
a. S A (k) = S A (j)
c. S A (E) = E
b. S A (D) = S B (D)
d. g is a line and S A (g) = g
e. If A β B and S A S B // g
7. Given A = (0,0) and B = (-4.1). Determine K such that the S A S B (K) = (6,2)
8. Given A = (-1,4), g = {(x, y) | y = 2x-1} and h = {(x, y) | y = -4x}
a. Determine the set equation SA (g) = g '.
b. Determine the set equation SA (h) = h '.
c. Determine the set equation SA (axis-x)
40. 36
d. Does the point (-5.6) lie on S A (g)? Explain!
9. Given circle C = {(x, y) | x 2
+ (Y-3) 2
= 4}, a line g = {(x, y) | y = x} and A = (3,2).
Show whether the point D = (2,5) is the element of the set M g S A (C).
10. Given line g, point P, and circle C, and suppose P β g does not intersect the C and P β
C. Moreover, C circle centered at A
a. By applying a halfturn, construct the line segment π΄πΜ Μ Μ Μ , so that the X β C, Y β g such
that the P midpoint of ππΜ Μ Μ Μ .
b. Prove that the construction is correct.
41. 37
CHAPTER VI
TRANSLATION
A. Vector
We have previously learned line segment in other lectures. In this part, we will
discuss about vector which is defined as follows
The notation of vector
- The notation of vector π΄π΅ is π΄π΅βββββ
- The notation of vector πΆπ· is πΆπ·βββββ
Vectorπ΄π΅ and πΆπ· are vector of which the points π΄ and πΆ are the the initial
points, and the points π΅ and π· are the terminal points.
As the illustration, look at the following figure:
The definition above can be interpreted also that π΄π΅βββββ = πΆπ·βββββ , if π π(πΆ) = π΅ and
π is the midpoint of π΄π·βββββ , as an illustration, look at the following figure:
Definition
Vector is a line segment of which one of the edge is the initial point and
the other edge is the terminal point.
Definition
π΄π΅βββββ is equivalent to πΆπ·βββββ that is notated as π΄π΅βββββ = πΆπ·βββββ , if there is a halfturn
π π( π΄) = π· where π is the midpoint π΅πΆβββββ
42. 38
The equivalence of two vectors satisfies a characteristic asserted by the
following theorem
Proof:
Look at the latest figure:
1. Suppose that π΄π΅βββββ = πΆπ·βββββ , if π is the midpoint of π΅πΆβββββ , then π π(π΄) = π·, by the
definition of equivalence. The diagonals of quadrilateral π΄π΅πΆπ· are bisected
in π. Then π΄π΅πΆπ· is a parallelogram.
2. Suppose that π΄π΅πΆπ· is a parallelogram, then diagonals π΄π· and π΅πΆ are
intersected in the midpoint π implying that π π(π΄) = π·. In other words, π is
the midpoint of both π΄π· and π΅πΆ.
So π΄π΅βββββ = πΆπ·βββββ
Consequence:
If π΄π΅βββββ = πΆπ·βββββ , then π΄π΅πΆπ·, π΄π΅βββββ and πΆπ·βββββ are either parallel or colinear
Theorem
If there are two vectors π΄π΅βββββ and πΆπ·βββββ which are not collinear then π΄π΅πΆπ· is
a parallelogram if and only if π΄π΅βββββ = πΆπ·βββββ
Theorem
Suppose π΄π΅βββββ , πΆπ·βββββ dan πΈπΉβββββ are vectors, then the following properties
apply:s
β’ π΄π΅βββββ = π΄π΅βββββ ( ππππππ₯ππ£π)
β’ If π΄π΅βββββ = πΆπ·βββββ then πΆπ·βββββ = π΄π΅βββββ (π π¦ππππ‘πππ)
β’ If π΄π΅βββββ = πΆπ·βββββ then πΆπ·βββββ = πΈπΉβββββ then π΄π΅βββββ = πΈπΉβββββ (π‘ππππ ππ‘ππ£π) are colinear.
Theorem
If π΄π΅βββββ is a vector and there is a point π, then there is a unique point Q
such that ππβββββ = π΄π΅βββββ .
43. 39
Proof :
Suppose R is the midpoint of π΅πβββββ , π = π π ( π΄) then π΄π΅βββββ = ππβββββ ππ ππβββββ = π΄π΅βββββ .
To prove the uniqueness of point, let π΄π΅βββββ = ππβββββ .
Then π π ( π΄) = π, because R is thee midpoint of π΅πβββββ and the image of π΄ by π π ,
then π = π meaning that ππβββββ is the only vector where π is the initial point and
π is the terminal point which is equivalent to π΄π΅βββββ
Corollary 1:
If π1( π₯1, π¦1), π2( π₯2, π¦2), and π3( π₯3, π¦3) are points with the specified
coordinates, then π(π₯3 + π₯2 β π₯1 , π¦3 + π¦2 β π¦1) is the only point that satisfies
π3 πβββββββ = π1 π2
ββββββββ
Corollary 2:
If ππ = ( π₯ π, π¦ π), π = 1,2,3,4 then π1 π2
ββββββββ = π3 π4
ββββββββ . If and only if π₯2 β π₯1 =
π₯4 β π₯3, and π¦2 β π¦1 = π¦4 β π¦3.
The last concept that is related to the vector is a scalar multiplication by
vector. Suppose that π΄π΅βββββ is vector and π is real number, so:
a. If π > 0, then ππ΄π΅βββββ is a vector π΄πβββββ which is defined as π΄πβββββ = π(π΄π΅βββββ ), π β
π΄π΅βββββ (ray π΄π΅).
b. If π < 0, then ππ΄π΅βββββ where π is the opposite ray of π΄π΅βββββ and π΄πβββββ = | π|(π΄π΅βββββββ )
44. 40
B. Translation
In this section, the concept of translation is introduced using the definition of
vector.
Proof.
Choose a coordinate system with π‘ as the π¦-axis and a line which is
perpendicular to π‘ as the π₯-axis, and s which its parallel to t.
Suppose π΄ = ( π1, π2) and π΅ = ( π1, π2).
If π ππ the midpoint of π΄β²β²π΅ββββββββ , then we have to prove π π( π΄) = π΅β²β²
.
If the equation of π is π₯ = π( π β 0), then
π΄β²β²
= ππ ππ‘( π΄) = ππ (βπ1, π2) = (2π + π1, π2)
π΅β²β²
= ππ ππ‘( π΅) = ππ (βπ1, π2) = (2π + π1, π2)
Because we know that π is the midpoint of π΄β²β²π΅ββββββββ , then
π = [
(2π + π1) + π1
2
,
π2 + π2
2
] , whereas
π π = [2 [
2π + π1 + π1
2
] β π1, 2 [
π2 + π2
2
] β π2] = (2π + π1, π2)
Evidently π π( π΄) = π΅β²β²
. Therefore π΄π΄β²β²ββββββββ = π΅π΅β²β²ββββββββ
Theorem
If π and π‘ are two parallel lines, then π΄ and π΅ are the two points, so π΄π΄β²β²ββββββββ =
π΅π΅β²β²ββββββββ with π΄β²β²
= ππ‘ π π (π΄) and π΅β²β²
= ππ‘ ππ (π΅).
45. 41
In the figure above, it is clear that every vector determines a translation.
Ifπ΄π΅βββββ is a vector then πΊ π΄π΅ is a symbol to address a translation in the length of π΄π΅.
Proof:
If π is an arbitrary point, it must be proved that πΊ π΄π΅( π) = πΊ πΆπ·( π). Suppose
πΊ π΄π΅( π) = π1 and πΊ πΆπ·( π) = π2, so ππ1 = π΄π΅βββββ and ππ2 = πΆπ·βββββ . Since π΄π΅βββββ = πΆπ·βββββ , ππ1
βββββββ =
ππ2
βββββββ meaning that π1 = π2 and πΊ π΄π΅ = πΊ πΆπ·
Proof:
Suppose π is an arbitrary point, if πβ = πΊ π΄π΅ (π) and πββ = ππ ππ‘ (π), so it is
necessary to prove that πβ = πββ
Definition
A mapping πΊ is a translation, if there is a vector π΄π΅βββββ such that for every point
π in a plane π has image πβ²
π€here πΊ( π) = πβ²
and ππβ²βββββββ = π΄π΅βββββ .
Theorem
If π΄π΅ = πΆπ· then πΊ π΄π΅ = πΊ πΆπ·
Theorem
Suppose π‘ and π as the two lines is parallel and πΆπ·βββββ is the vector that
perpendicular to π and π‘, with πΆ β π and π· β π . If π΄π΅βββββ = 2πΆπ·βββββ so πΊ π΄π΅ = ππ ππ‘
46. 42
As the property of translation suggests, if πΊ π΄π΅(π) = πβ then ππβββββββ = π΄π΅βββββ . Since,
π΄π΅βββββ = 2πΆπ·βββββ , ππβββββββ = 2πΆπ·βββββ .
Related to πΆββ = ππ ππ‘( πΆ), πΆ β π‘ so πΆββ = ππ‘(πΆ). It means that π· is the
midpoint of πΆπΆβββββββββ implying that πΆπΆβββββββββ = 2 πΆπ·βββββ . Since πΆπΆβββββββββ = ππβββββββββ , ππβββββββββ = 2 πΆπ·βββββ = ππβββββββ ,
and it means that πβ = πββ, so πΊ π΄π΅ = ππ ππ‘.
Notes
1) Each translation πΊ π΄π΅ can be written as the composition between two reflections
in two lines which are perpendicular to π΄π΅Μ Μ Μ Μ and it is Β½ π΄π΅ in length.
2) If π΄π΅ is a line and πΆ is the midpoint of π΄π΅Μ Μ Μ Μ whereas π‘, π and π are the perpendicular
lines to π΄π΅ in π΄, πΆ and π΅ respectively, then πΊ ππ = ππ ππ‘ = π π ππ
3) Since any translation can be written as a composition of two reflections, whereas
a reflection is a transformation is isometry then a translation is an isometry
transformation.
Translation is a direct isometry
Proof:
According to latest figure, πΊ π΄π΅ = ππ ππ‘ = π π ππ
While πΊ π΅π΄ = ππ‘ ππ = ππ π π
( πΊ π΄π΅)β1
= ( ππ ππ‘)β1
= ππ‘
β1
. ππ β1
= ππ‘ ππ = πΊ π΅π΄
Theorem
If πΊ π΄π΅ is a translation, then ( πΊ π΄π΅)β1
= πΊ π΅π΄
47. 43
So, ( πΊ π΄π΅)β1
= πΊ π΅π΄.
C. The Closeness of translation
In the previous section, it is explained that a translation can be expressed in the
form of a composition of two reflections. This section will describe that the composition
of the two translations is a translation as well. The assertion refer to the following
theorem:
Proof :
Suppose that π = πΆπ·βββββ , π β₯ π ππ πΆ, π β₯ π ππ π·
π΄π΅ is a vector from π to π, therefore π΄π΅βββββ = 2πΆπ·βββββ , so πΊ π΄π΅ = π π· π πΆ. While π π· =
π π ππ and ππ = ππ ππ.
Then π π· π πΆ = (π π ππ)(ππ ππ) = π π(ππ ππ)ππ = π π ππ. ππ πΊ π΄π΅ = π π· π πΆ.
Example 1
Given π΄ = (3, β1), π΅ = (1,7) and πΆ = (4,2). Find the coordinate of a point
π· such that πΊ π΄π΅ = π π· π πΆ.
Solution:
Suppose that πΈ is a point such that πΆπΈ = π΄π΅, then πΈ[4 + (1 β 3),2 + (7 β (β1))]
or
πΈ = (2,10). If π· is the midpoint πΆπΈΜ Μ Μ Μ , then π· = (3,6), implying that πΆπΈβββββ = 2πΆπ·βββββ .
Theorem
If πΊ π΄π΅ is a translation, C and D are points such that π΄π΅βββββ = πΆπ·βββββ , then πΊ π΄π΅ = π π π πΆ.
48. 44
Thus, π΄π΅βββββ = 2πΆπ·βββββ , which obtains πΊ π΄π΅ = π π· π πΆ where π· (3,6).
Proof:
Suppose that πΊ π΄π΅ is a translation and πΆ is any point and suppose that πΈ is also a point
so
πΆπΈβββββ = π΄π΅βββββ . If π· mid point πΆπΈΜ Μ Μ Μ Μ then πΆπΈβββββ = 2πΆπ·βββββ .
According to previous theorem πΊ π΄π΅ = π π· π πΆ, then πΊ π΄π΅ π πΆ = π π· π πΆ π πΆ = π π· πΌ =
π π·. So πΊ π΄π΅ π πΆ = π π·.
As a result of the theorem above is:
If ππ΄, π π΅ and π πΆ is half round, then ππ΄ π π΅ π πΆ = π π·, with π· is points that satisfy π΄π·βββββ =
π΅πΆβββββ .
To declare the composition of two translations is a translation in Cartesian
coordinates, consider the following theorem.
Proof:
For π(π₯, π¦), π(π) = (π₯ + π, π¦ + π). Suppose that πβ = πΊ ππ΄(π) then ππβ = ππ΄
implying that πβ = (π₯ + π β 0, π¦ + π β 0) = (π₯ + π, π¦ + π). Thus π(π) =
πΊ ππ΄(π) for each π β π. In other words πΊ ππ΄ = π.
Example 2
Suppose that πΊ π΄π΅ is a translation which takes point π΄(2,3) to point π΅(4,1) and πΊ πΆπ·
is a translation which takes point πΆ(β3,4) to point π΅(0,3). If π(π₯, π¦). Determine the
coodinate of πΊ πΆπ· πΊ π΄π΅(π).
Theorem
The composition of a translation and a halfturn is a halfturn.
Theorem
If πΊ ππ΄ is a translation where the coordinates of point 0 and point π΄ are respectively
(0,0) and (π, π). π is a transformation that maps each point π (π₯, π¦) to π (π) =
(π₯ + π, π¦ + π) then πΊ ππ΄ = π.
49. 45
Solution:
Suppose that 0β = πΊ π΄π΅(0) and 0ββ = πΊ πΆπ·(π) then ππβ²βββββββ = π΄π΅βββββ and ππβ²β²ββββββββ = πΆπ·βββββ .
Then 0β = (0 + 4 β 2,0 + 1 β 3) = (2, β2) and
0ββ = (0 + 0 + 3,0 + 3 β 4) = (3, β1).
So πΊ π΄π΅(π) = (π₯ + 2, π¦ β 2) and πΊ πΆπ·(π) = (π₯ + 3, π¦ β 1).
Thus πΊ πΆπ· πΊ π΄π΅( π) = πΊ πΆπ·[( π₯ Β± 2, π¦ β 2)]
= ( π₯ + 2 + 3, π¦ β 2 β 1)
= (π₯ + 5, π¦ β 3)
Exercises
1. Suppose there are 4 points that are wirtten as π΄, π΅, πΆ and π· of which each pair of
three points is not collinear. Skecth the following:
a. Point E such that πΆπΈβββββ = π΄π΅βββββ
b. Point F such that π·πΉβββββ = π΅π΄βββββ
c. ππ΄(π΄π΅)
2. Given points π΄(0,0), π΅(5,3), and πΆ(β2,4). Find the coordinate :
a. R so that such that π΄π βββββ = π΅πΆβββββ
b. S so that such that πΆπββββ = π΄π΅βββββ
c. T so that such that ππ΅βββββ = π΄πΆβββββ
3. If π΄(1,3), π΅(2,7), dan πΆ(β1,4) are the vertices of parallelogram π΄π΅πΆπ·. Determine
the coordinate of the point π·.
4. Suppose there are two lines π and β whcih are parallel, point π β π, and point π
neither on π nor on β.
a. Sketch πβ = πβ ππ(π) and πβ = ππ πβ(π)
b. Prove that ππβ²ββββββ = ππβ²βββββββ
5. Given a line π and circles πΏ1 and πΏ2. Line π are not cuting the circles πΏ1 and πΏ2. Use
a transformation to draw a square having vertices located on π, a vertex that is located
on πΏ1and the other vertex points located on πΏ2
50. 46
6. If π0 = (0,0), π1 = (π₯1, π¦1), π2 = (π₯2, π¦2), and π3 = (π₯3, π¦3) while π > 0.
a. Find the coordinate of π such that π0 π1
ββββββββ = ππ0 π1
ββββββββ
b. Find the coordinate of π such that π1 πβββββββ = ππ1 π2
ββββββββ
c. Is the statement βIf π3 πβββββββ = ππ1 π2
ββββββββ , then π = {π₯3 + π(π₯2 β π₯1), π¦3 + π(π¦2 β
π¦1)}.β valid for π < 0?
7. Given π΄, π΅, and πΆ which are not collinear. Sketch:
a. πΊ π΄π΅(π΄) and πΊ π΄π΅(π΅).
b. πΊ π΄π΅(πΆ).
c. The lines π and β where π΄ β π and πΊ π΄π΅ = πβ ππ.
8. Suppose there are two points that notated as π΄ and π΅ and line π such that π β₯ π΄π΅βββββ .
Sketch:
a. Line β such that πβ ππ = πΊ π΄π΅.
b. Line π such that ππ ππ = πΊ π΄π΅.
c. Line π such that πβ = πΊ π΄π΅(π).
d. Point πΆ such that the πΊ π΅π΄ (πΆ) = π΅
9. Suppose lines π and β are parallel and there is point π΄ which is not on the line.
a. Draw the point π΅ such that πβ ππ = πΊ π΄π΅
b. Draw the point πΆ such that ππ πβ = πΊ2π΄πΆ
10. Given π΄ (2,3) and π΅ (β4,7), determine the equation of a line π and β such that
πβ ππ = πΊ π΄π΅
11. Given three points π΄ (β1,3), π΅ (5, β1) and πΆ (2,4)
a. Find the coordinate πΆ β² = πΊ π΄π΅ (πΆ)
b. Find the equation of lines π and β so πΆ β π and so πβ ππ = πΊ π΄π΅
51. 47
12. The edges of a river are depicted with two parallel lines that is written as π‘ and π
(see figure). Above the river, there will be built a bridge and according to a good
construction, the bridge must be made perpendicular to the direction of the river.
Where is the bridge should be constructed usch that the distance fromthe town π· to
the town πΈ will be as short as possible.
52. 48
CHAPTER VII
ROTATION
A. Directed Angle
Angle has been introduced previously as the combination between two rays that have
identical initial point. For example, angle ABC notated by β ABC is formed by BA and BG
rays. We can see angle ABC in the following figure:
Both figures (a and b) certainly illustrate angle ABC. But for the next discussion both
figures will be distinguished. It is distinguished by using initial ray and terminal ray of an
angle.
Itβs used to determine what type of positive angle or negative angle from an angle.
Such angle is called directed angle
To symbolize an angle, for example β ABC is a directed angle where π΅π΄βββββ rays is the
initial ray and π΅πΆβββββ is the terminal ray notated as β ABC
For another purpose, <ABC cannot be written as <BCA. For directed angle <CBA, the
initial ray is π΅πΆβββββ and the end ray is π΅π΄βββββ .
In Euclidean geometry, it has been studied about the magnitude of an angle, i.e. every
angle ABC has magnitude in the interval from 0Β°
to 180Β°
, which is notated by 0Β°
β€ π <
π΄π΅πΆ β€ 180Β°
.
It is the same as a directed angle < π΄π΅πΆ which always has a magnitude, however, it has
following properties:
a) If the triple orientation (π΅πΆπ΄) is positive, then π < π΄π΅πΆ = π < π΄π΅πΆ
b) If the triple orientation (π΅πΆπ΄) is negative, then π < π΄π΅πΆ = β(π < π΄π΅πΆ)
Definition
Directed angle is an angle of which one of the ray is the initial ray and the other
is the terminal ray
53. 49
When < π΄π΅πΆ is an angle then < π΄π΅πΆ = < πΆπ΅π΄, so π < π΄π΅πΆ = βπ < πΆπ΅π΄.
But for a directed angle < π΄π΅πΆ it applies for the π < π΄π΅πΆ =
βπ πΆπ΅π΄ since the orientation of the π΅π΄πΆ is always contrary to the orientation of the π΅πΆπ΄.
If there are two intersecting lines not perpendicular to each other, then its angle is the
acute angle.
In the figure, the magnitude of the angle between line g and k is 80 Β°, and the
magnitude of the angle between h and k is -60 Β°.
The angle between two lines can be described as follows:
Suppose that g and h intersect at point O, point P is on line g, points Q and R are located
on line h as suggested in the following figure:
If β POR is an acute angle, then the magnitude of line g to line h is equal to β POR,
whereas if the angle β POR is an obtuse angle, then the angle from line g to line h equals the
magnitude of β POQ. Suppose mβ POR = 140 Β°, then the magnitude from line g to line h is
mβ POQ = 40 Β°, while the magnitude of line h to line g is mβ QOP = -40 Β°.
B. Rotation
It will be discussed in the present section, the composition of the two reflection in the
lines that are not parallel and in intersecting lines but are not perpendicular to each other.
The composition of the two reflections will produce an isometry either in the form of a
rotation and a translation. The composition is a basic theorem of rotation.
54. 50
Proof:
Case 1 : Pβ s and Qβ s
Aβ= MtMs(A) means that Aβ=A. Pβ=MtMs(P) and Qβ=MtMs(Q). Sincethe points A, P and
Q are located on are collinear, Aβ, Pβ, and Qβ are collinear. Thus, lines PPβ and QQβ
intersect at point A. Thus m (< PAPβ) = m (< QAQβ).
Case 2 : Pβ s and Qβ s
m ( < PAPβ) = m ( < PAQ ) + m ( < QAPβ), then m (< QAQβ) = m (< QAPβ) + m (<
PβAQβ), since m (< PAQ)= m (< PβAPβ) so m (< PAPβ) = m (< QAQβ).
Case 3 : Pβ s and Qβ s
Theorem B
If s and t intersect at point A but are not perpendicular, the points P and Q are the
points that are different from A, then m (< PAPβ) = m (< QAQβ) with Pβ =MtMs(P)
and Qβ =MtMs(Q)
55. 51
RA, Ξ±(P)
PA
RA, -Ξ±(P)
In the case 3, we can prove that if Pβ= MtMs(P) and Qβ=MtMs(Q), so m (< QAQβ = m (<
PAPβ).
Therefore, by transformation of MtMs, every point rotates by the same directed angle,
rotating the same point. Such process is called a rotation.
Based on the definition, the rotation π π΄,πΌ, just has an invarioant point, i.e. A (the
center of the rotation). The image of point the P by rotation π π΄,πΌ, is a point of a circle
where A is the center and AP is the radius.
Since magnitude Ξ± of the angle rotation is between -1800
and 1800
, so Ξ±>0, if the
direction of the angle is opposite to the clockwise direction, and Ξ±<0 if the direction of the
angle is the same as the of the clockwise direction.
Defenition
If A is a point and πΌ is angle where -180Β°<πΌ<180Β°, a rotation with center A and
angle πΌ denoted by π π΄,πΌ is a function from V to V defined by
1. If P = A, so π π΄,πΌ(P) = P.
2. If P β A, so π π΄,πΌ(P) = Pβ, then m (< PAPβ) = πΌand APβ=AP.
Theorem
If s and t are two lines that intersect on A, s and t arenβt perpendicular, and if the
magnitude of the angle from line s to t is Ξ±/2 , then RA,Ξ± =MtMs
56. 52
A P
u
s
t
v
Β½ Ξ±
Β½ Ξ±
K
t
s
Β½ Ξ±
Proof:
Suppose that a point K β A located on s. If Kβ = MtMs(K) so m(<KAKβ) = 2. Β½ Ξ± = Ξ±.
Since <KAKβ=Ξ±, so RA,Ξ± =MtMs.
Example
If RA,Ξ± is a rotation that maps point P to Pβ, determine two pairs of lines which can be
used as some reflection axes such that the composition of these reflections is a rotation.
Solution
1. Suppose that s = π΄πΜ Μ Μ Μ , t is the bisecting line of <PAPβ, and suppose the magnitude of
angle from s to t is Β½ Ξ±, so RA,Ξ± = MtMs.
K
A
Theorem
1. Rotation with A as the center and the magnitude Ξ± (RA,Ξ±) is a transformation.
2. Every rotation is direct isometry
3. The composition of two reflections is either rotation or translation
57. 53
2. Suppose u = π΄πβ²Μ Μ Μ Μ Μ , and v is a line passing through A, then the angle magnitude from u to
v is Β½ Ξ±, so RA,Ξ± = MvMu.
C. Rotation Composition
We have proved that the rotation is a transformation. Because the composition of two
transformations is transformation, what about the composition of two rotations? Whether it
is a rotation or another transformation? For the composition of two transformations, the
following matters are discussed:
a. If the centers of the rotations are the same
b. If the centers of rotation are different
If the centers of the rotations are the same, the composition of two rotations is a
rotation with the same center and the magnitude of the angle is the sum of the angles
(provided that if the sum is greater than 180Β° it must be substracted by 360Β°while if the
sum is less than -180Β° it must be added by 360Β°).
Meanwhile, the composition of the rotations having different centers, can be in the
form of a rotation of which the center is different to the centers of th composed rotations .
In addition, the angle magnitude is the sum of the angles of the rotations following the
previous case order. If the sum of the angles of is zero, then the composition of two
rotations forms a translation.
In general it can be conluded as follows:
If RA, πΌ1 andRB, πΌ2 is two rotation, then RB, πΌ2RA, πΌ1 = Rc, πΌ with the conditions of πΌ as
follows:
1. If 0Β° < | πΌ1 + πΌ2| < 180Β° π‘βππ πΌ = πΌ1 + πΌ2.
2. If πΌ1 + πΌ2 > 180Β° π‘βππ πΌ = ( πΌ1 + πΌ2) β 360Β°.
3. If πΌ1 + πΌ2 < β180Β° π‘βππ πΌ = ( πΌ1 + πΌ2) + 360Β°.
Example
RA,120 . RA,30= RA,150
RB,160 . RB,40= RB,-160 .
RC,-150 .RC,-50 .= RC,160 .
While ifπΌ1 + πΌ2 = 0Β°, π‘βππ RB, πΌ2RA, πΌ1is a rotation .
Theorem
The composition of two rotations is a rotation or translation.
58. 54
Proof :
Assume there are rotations π π΄,πΌand π π΅,π½. Draw a line s = π΄π΅β‘ββββ , if m(β XAY) = m(β XAZ)
=
1
2
πΌ, then π π΄,πΌ = ππ ππ‘ and π π΅,π½ = π π’ ππ .
So π π΅,π½ π π΄,πΌ = π π’ ππ ππ ππ‘= π π’ ππ‘.
If u is parallel to t then π π΅,π½ π π΄,πΌ is a translation, and if u and t intersect at C, then
π π’ ππ‘ is a rotation centered at C. Assume π πΆ,0 = π π΅,π½ π π΄,πΌthen in πΌ, π½, dan π, there is
following relationship:
m(β ABC) =
1
2
πΌ, m(β BAC) =
1
2
π½.
Therefore m(β PCB) =
1
2
πΌ +
1
2
π½. It means that the angle from t to u is
1
2
πΌ +
1
2
π½, so that
π = πΌ + π½.
If πΌ + π½ > 180Β°, then π= ( πΌ + π½) β 360Β°.
EXERCISES
1. Given points A and P are different. Construct the following
A. RA,90 (P)
B. RA,150 (P)
C. RA,45 (P)
D. Q such that RA,30 (Q) = P
2. Given m (<ABC) = 40Λ and m (<BAD) = 120Λ, determine:
A. m (<DAB),m(BCA), m(<ECA)
B. The magnitude of angles fom
π΄π΅
β to
π΅πΆ
β , from
π΄πΆ
β to
π΅πΆ
β , and from
π΄π΅
β to
π΄πΆ
β
3. Suppose point A and P are different, find m(<PAP') if Pβ
is the image P by the
following transformations:
a. RA,30, RA,90
b. RA,-60, RA,120
c. RA,135, RA,90
d. RA,-120 RA,-150
4. Simplify the following transformation compositions:
A. RA,30 RA,60
59. 55
B. RA,120 RA,-90
C. RA,135 RA,-90
D. RA,-60 RA,45
E. RA,-120 RA,-150
F. RA,-120 RA,90
5. Suppose there are two lines s and t which intersect at point A and two points P and Q
not on the lines.
A. Construct point Pβ
= MtMs(P)
B. Construct point Qβ
= MsMt(Q)
C. Construct point Pβ
= MtMs(P)
D. if m (<PAPβ
) = 118Λ what is the angle magnitude between s and t.
6. Given points A, B and B 'such that RA, Ξ± (B) = B'. Construct two lines s and t such that
MsMt = RA, Ξ±
7. Given 0 is the center point of the orthogonal coordinate system and A=(1,0). Find the
coordinates of the following points:
a. π 0,90( π΄)
b. π 0,45(π΄)
c. π 0,120( π΄)
d. π 0,β135( π΄)
8. Writen the equation of lines s and t such that ππ ππ‘ equals the following rotation, if
A=(1,3) and 0 is the center point.
a. π π,90
b. π π,β180
c. π π,120
d. π π΄,90
e. π π΄,β90
9. If A is the center point of an orthogonal coordinate system and π = {(π₯, π¦)|π¦ = 2π₯ β
3} find the equation of sβ
= π π΄,90( π)
10. If I is a circle with radius 2 and the center in A=(β2, β2) and if B = (0,0), find the
equation of Iβ=π π΄,90( πΌ)