The document discusses division of intervals in coordinate geometry. It states that for a 2 unit math exam, interval division questions are restricted to finding the midpoint, which divides an interval in a 1:1 ratio. For Extension 1 exams, intervals can be divided in any ratio internally or externally. The midpoint formula averages the x- and y-coordinates of two points to find the midpoint, while the distance formula calculates the length of the hypotenuse between two points using Pythagoras' theorem.
11 X1 T05 01 division of an interval (2010)Nigel Simmons
Here are the steps to solve this problem:
1) Write down the endpoints of the interval: (-3,4) and (5,6)
2) The ratio given is 1:3
3) Let the interval from the first endpoint to P be 1 unit
4) Then the interval from P to the second endpoint is 3 units
5) Use the ratio of distances formula:
xP - x1 / x2 - x1 = 1/(1+3)
yP - y1 / y2 - y1 = 1/(1+3)
6) Solve the two equations simultaneously to find the coordinates of P.
The document discusses coordinate geometry concepts including the distance formula and midpoint formula. It explains that the distance formula calculates the length of the hypotenuse between two points using Pythagoras' theorem. The midpoint formula averages the x- and y-coordinates of two points to find the midpoint. It also discusses dividing intervals, noting that for a level 2 math exam it is restricted to midpoint divisions in a 1:1 ratio, while an extension 1 exam can involve any ratio for internal or external divisions. Examples are provided to illustrate the concepts.
1. The document provides examples of using the distance and midpoint formulas to calculate the distance between two points and find the midpoint between two points on a coordinate plane.
2. It gives the formulas for distance (d=(x2-x1)2+(y2-y1)2) and midpoint ((x1+x2)/2,(y1+y2)/2) and works through examples of using each.
3. It also includes a word problem example about meeting a friend in Washington D.C. where the midpoint formula is used to determine which landmark is closest to meet at.
This document contains a math lesson on the midpoint formula. It begins with examples of using the midpoint formula to find the coordinates of the midpoint of a line segment given the coordinates of the endpoints. It then provides practice problems for students to find midpoints and missing endpoint coordinates. The document aims to teach students how to use the midpoint formula to solve geometry problems.
The document discusses two approaches to calculating the distance between two points: the traditional distance formula approach and an alternative approach using a drawing and the Pythagorean theorem. It provides an example of finding the distance between points (-3,4) and (5,-3) and step-by-step works through the alternative approach of drawing a right triangle between the points and applying the Pythagorean theorem to calculate the distance as the hypotenuse of 10.6 units. The document suggests asking students to estimate distances, draw diagrams, calculate, and reflect on their work.
1. The document provides instructions for an assignment that is due on Monday May 7th. It includes completing lesson 10.6 questions 2-4, 10-28, 38-42 (evens), and checkpoint 10.6. There is also a final exam scheduled for Thursday May 10th.
2. Lesson 10.6 covers the distance and midpoint formulas for finding the distance and midpoint between two points on a coordinate plane. It provides examples of using the formulas to find the distance between points and find the midpoint of segments.
The document is a mathematics exam paper for Form Four students in Negeri Sembilan, Malaysia. It consists of 40 multiple choice questions testing various mathematical concepts. The questions cover topics like rounding, standard form, algebraic expressions, sets, probability, geometry, trigonometry, and more. The document provides mathematical formulas that may be helpful in answering the questions.
The document discusses distance and midpoint formulas. It provides the formulas for calculating the distance between two points and finding the midpoint between two points. Examples are included to demonstrate using the formulas to find the distance and midpoint for pairs of points in a coordinate plane. The key formulas are:
Distance Formula: d = √(x2 - x1)2 + (y2 - y1)2
Midpoint Formula: Midpoint = (x1 + x2)/2, (y1 + y2)/2
11 X1 T05 01 division of an interval (2010)Nigel Simmons
Here are the steps to solve this problem:
1) Write down the endpoints of the interval: (-3,4) and (5,6)
2) The ratio given is 1:3
3) Let the interval from the first endpoint to P be 1 unit
4) Then the interval from P to the second endpoint is 3 units
5) Use the ratio of distances formula:
xP - x1 / x2 - x1 = 1/(1+3)
yP - y1 / y2 - y1 = 1/(1+3)
6) Solve the two equations simultaneously to find the coordinates of P.
The document discusses coordinate geometry concepts including the distance formula and midpoint formula. It explains that the distance formula calculates the length of the hypotenuse between two points using Pythagoras' theorem. The midpoint formula averages the x- and y-coordinates of two points to find the midpoint. It also discusses dividing intervals, noting that for a level 2 math exam it is restricted to midpoint divisions in a 1:1 ratio, while an extension 1 exam can involve any ratio for internal or external divisions. Examples are provided to illustrate the concepts.
1. The document provides examples of using the distance and midpoint formulas to calculate the distance between two points and find the midpoint between two points on a coordinate plane.
2. It gives the formulas for distance (d=(x2-x1)2+(y2-y1)2) and midpoint ((x1+x2)/2,(y1+y2)/2) and works through examples of using each.
3. It also includes a word problem example about meeting a friend in Washington D.C. where the midpoint formula is used to determine which landmark is closest to meet at.
This document contains a math lesson on the midpoint formula. It begins with examples of using the midpoint formula to find the coordinates of the midpoint of a line segment given the coordinates of the endpoints. It then provides practice problems for students to find midpoints and missing endpoint coordinates. The document aims to teach students how to use the midpoint formula to solve geometry problems.
The document discusses two approaches to calculating the distance between two points: the traditional distance formula approach and an alternative approach using a drawing and the Pythagorean theorem. It provides an example of finding the distance between points (-3,4) and (5,-3) and step-by-step works through the alternative approach of drawing a right triangle between the points and applying the Pythagorean theorem to calculate the distance as the hypotenuse of 10.6 units. The document suggests asking students to estimate distances, draw diagrams, calculate, and reflect on their work.
1. The document provides instructions for an assignment that is due on Monday May 7th. It includes completing lesson 10.6 questions 2-4, 10-28, 38-42 (evens), and checkpoint 10.6. There is also a final exam scheduled for Thursday May 10th.
2. Lesson 10.6 covers the distance and midpoint formulas for finding the distance and midpoint between two points on a coordinate plane. It provides examples of using the formulas to find the distance between points and find the midpoint of segments.
The document is a mathematics exam paper for Form Four students in Negeri Sembilan, Malaysia. It consists of 40 multiple choice questions testing various mathematical concepts. The questions cover topics like rounding, standard form, algebraic expressions, sets, probability, geometry, trigonometry, and more. The document provides mathematical formulas that may be helpful in answering the questions.
The document discusses distance and midpoint formulas. It provides the formulas for calculating the distance between two points and finding the midpoint between two points. Examples are included to demonstrate using the formulas to find the distance and midpoint for pairs of points in a coordinate plane. The key formulas are:
Distance Formula: d = √(x2 - x1)2 + (y2 - y1)2
Midpoint Formula: Midpoint = (x1 + x2)/2, (y1 + y2)/2
The document explains the distance formula and how to calculate the distance between two points in a coordinate plane. It provides the formula for distance as the square root of the sum of the squares of the differences between the x-coordinates and y-coordinates of the two points. Several examples are given of using the distance formula to calculate the distance between points with given coordinates.
Find the distance between two points
Find the midpoint between two points
Find the coordinates of a point a fractional distance from one end of a segment
This document discusses formulas for finding the distance and midpoint between two points. It provides the distance formula, midpoint formula, and worked examples finding the distance and midpoint between points like (-3, 1) and (2, 3). It also includes a bonus problem asking to find all possible values of a given the distance between (4, a) and (1, 6) is 5 units. An alternate method for finding distance by graphing the points and using the lengths of the sides of the right triangle formed is also described.
The document discusses determining the inverse of matrices. It provides examples of finding the inverse of 2x2 matrices using simultaneous linear equations and formulas. The simultaneous equation method involves setting up equations where the product of the matrix and its inverse equals the identity matrix. The formula method uses the determinant of the matrix and the positions of the elements to determine the inverse. Exercises provide additional practice finding inverses of matrices using these methods.
1) The document discusses the distance and midpoint formulas. The distance formula calculates the distance between two points on a coordinate plane using their x and y coordinates, while the midpoint formula calculates the midpoint between two points.
2) An example uses the distance formula to calculate the distance between points (2, -6) and (-3, 6), finding it to be 13.
3) Another example finds the midpoint between points (1, 2) and (-5, 6) to be (-2, 8) using the midpoint formula.
These slides about functions between two sets. Topics included are
1- The definition of a function as relations between two sets
2- Examples of infinite functions
3- Relations which are not functions
4- Domain and range of real functions
5- Identical functions
Videos explaining these slides are available in the following links
1- The definition of a function as relations between two sets
https://youtu.be/Vi3N2vLySd0
2- Examples of infinite functions
https://youtu.be/Fex82-Ml55c
3- Relations which are not functions
https://youtu.be/abhbALKcHn8
4- Domain and range of real functions
https://youtu.be/82LJ5MXAKRQ
5- Identical functions
https://youtu.be/ZOIt5JxoBxo
The reference book for these slides is
A Transition to Advanced Mathematics 8th Edition,
by Douglas Smith, Maurice Eggen, Richard St. Andre. ISBN-13: 978-1285463261, published by Cengage Learning (August 6, 2014).
https://www.cengagebrain.co.uk/shop/i...
The document discusses the distance formula and how to calculate the distance between two points. It provides the formula: Distance = √(x2 - x1)2 + (y2 - y1)2. Several examples are shown of using the distance formula to find the distance between points. The document also covers finding the midpoint between two points using the formula: Midpoint = (x1 + x2)/2, (y1 + y2)/2.
The document discusses the distance formula for finding the distance between two points P and Q on a line. It states that the distance is equal to the absolute value of P - Q. It provides an example where P = -2 and Q = 3, and calculates the distance as 5 using the formula |P - Q|. It then explains that we can square the difference P - Q to avoid issues with the absolute value notation, so the distance formula is the square root of (P - Q)2.
The document explains how to calculate the distance between two points using the distance formula. It shows that if points are horizontal or vertical, you can use their x- or y-coordinates alone, but otherwise you need to use the Pythagorean theorem to form a right triangle and find the hypotenuse. The distance formula is given as the square root of the sum of the squared differences between corresponding x- and y-coordinates of the two points. An example using points (3, -5) and (-1, 4) demonstrates applying the formula to find the distance between two points.
These slides cover the relation between two sets, domain and range of a relation, the inverse of a relation, and composition of relations.
Videos explaining these slides are available
Part1: Definition of relations https://youtu.be/MR0ALeKvaSc
Part2: Domain and range of a relation https://youtu.be/kBlUWTDn-Z4
Part3: The inverse of a relation https://youtu.be/Uv5KFusvvsY
Part4: Composition of relations https://youtu.be/IvVIjJqqPH0
Reference for these slides are
A Transition to Advanced Mathematics 8th Edition,
by Douglas Smith, Maurice Eggen, Richard St. Andre. ISBN-13: 978-1285463261, published by Cengage Learning (August 6, 2014).
https://www.cengagebrain.co.uk/shop/i...
This document provides formulas and examples for calculating the distance and midpoint between two points.
The distance formula is used to find the distance between two points (x1, y1) and (x2, y2). The midpoint formula finds the midpoint between two points by taking the average of the x-coordinates and y-coordinates.
Several examples demonstrate using the distance and midpoint formulas to find distances, midpoints, and missing endpoint coordinates. The key steps are to work inside parentheses, do exponents before addition, and plug points into the formulas.
1. The document discusses the distance and midpoint formulas in coordinate geometry.
2. It provides the formulas for finding the distance between two points and the midpoint of a line segment.
3. Examples are given of using the formulas to find distances and midpoints, as well as classifying triangles and finding endpoint coordinates given a midpoint.
1. The document discusses finding the midpoint and distance between points in a coordinate plane. It provides examples and steps for using the midpoint formula and distance formula.
2. The midpoint formula is used to find the midpoint between two points by taking the average of the x-coordinates and y-coordinates. The distance formula and Pythagorean theorem can both be used to find the distance between points.
3. Examples include finding midpoints and distances between points, as well as applications to sports such as finding the distance of a throw between bases in a baseball diamond.
1) The document discusses distance and midpoints between two points in coordinate geometry. It defines the distance formula as the absolute value of x2 - x1 plus y2 - y1.
2) It presents the midpoint formulas, where the x-coordinate of the midpoint is (x1 + x2)/2 and the y-coordinate is (y1 + y2)/2.
3) Examples are given to find the distance between two points, and to determine the midpoint or endpoints of a line segment given coordinates of two points.
This document introduces the distance formula, which is used to calculate the distance between two points (x1, y1) and (x2, y2) on a coordinate plane. The distance formula is the square root of (x1 - x2) squared plus (y1 - y2) squared. Several examples are worked through to demonstrate finding the distance between points using their coordinates. Practice problems are also provided for the reader to work through on their own.
The document discusses two methods for simplifying complex fractions. A complex fraction is a fraction with fractions in the numerator or denominator. The first method reduces the complex fraction to an "easy" problem by combining the numerator and denominator terms into single fractions. The second method multiplies the lowest common denominator of all terms to both the numerator and denominator of the complex fraction. An example using each method is provided to demonstrate the simplification process.
This an animated slides for students. Introduce basis concept of proofs to students. Direct proofs. Please search for slides Proof methods-teachers. If you want to teach using these slides.
The document discusses using the distance formula to calculate the distance between two points on a coordinate plane. It gives the formula for distance as the square root of (x1 - x2) squared plus (y1 - y2) squared, where (x1, y1) and (x2, y2) are the coordinates of the two points. It provides an example of using the distance formula and exercises for the reader to practice calculating distances between points.
The document discusses multiplying and dividing variable expressions. It provides examples of simplifying expressions using the distributive property and the property of the opposite of a sum. It also demonstrates dividing variable expressions by writing the division as a fraction and simplifying. Key steps include distributing terms, dividing each term in the numerator by the denominator, and evaluating expressions for given variable values.
11 x1 t11 06 tangents & normals ii (2013)Nigel Simmons
This document discusses finding tangents and normals to parabolas defined by equations of the form y = ax^2 using Cartesian coordinates. It shows that the slope of a tangent line at a point (x1, y1) is 2x1/a, and the slope of the corresponding normal is -a/x1. It also describes how to determine whether a given line will cut, touch, or miss the parabola based on the discriminant of the equation for their intersection.
The document discusses tests for determining if triangles are similar. There are three tests: (1) corresponding sides are in proportion (SSS), (2) two pairs of corresponding sides are in proportion and included angles are equal (SAS), (3) all three angles are equal (AA). An example problem finds the length of side AD using the properties of similar triangles. The side lengths are in the same ratio as the corresponding sides between the two triangles.
The document explains the distance formula and how to calculate the distance between two points in a coordinate plane. It provides the formula for distance as the square root of the sum of the squares of the differences between the x-coordinates and y-coordinates of the two points. Several examples are given of using the distance formula to calculate the distance between points with given coordinates.
Find the distance between two points
Find the midpoint between two points
Find the coordinates of a point a fractional distance from one end of a segment
This document discusses formulas for finding the distance and midpoint between two points. It provides the distance formula, midpoint formula, and worked examples finding the distance and midpoint between points like (-3, 1) and (2, 3). It also includes a bonus problem asking to find all possible values of a given the distance between (4, a) and (1, 6) is 5 units. An alternate method for finding distance by graphing the points and using the lengths of the sides of the right triangle formed is also described.
The document discusses determining the inverse of matrices. It provides examples of finding the inverse of 2x2 matrices using simultaneous linear equations and formulas. The simultaneous equation method involves setting up equations where the product of the matrix and its inverse equals the identity matrix. The formula method uses the determinant of the matrix and the positions of the elements to determine the inverse. Exercises provide additional practice finding inverses of matrices using these methods.
1) The document discusses the distance and midpoint formulas. The distance formula calculates the distance between two points on a coordinate plane using their x and y coordinates, while the midpoint formula calculates the midpoint between two points.
2) An example uses the distance formula to calculate the distance between points (2, -6) and (-3, 6), finding it to be 13.
3) Another example finds the midpoint between points (1, 2) and (-5, 6) to be (-2, 8) using the midpoint formula.
These slides about functions between two sets. Topics included are
1- The definition of a function as relations between two sets
2- Examples of infinite functions
3- Relations which are not functions
4- Domain and range of real functions
5- Identical functions
Videos explaining these slides are available in the following links
1- The definition of a function as relations between two sets
https://youtu.be/Vi3N2vLySd0
2- Examples of infinite functions
https://youtu.be/Fex82-Ml55c
3- Relations which are not functions
https://youtu.be/abhbALKcHn8
4- Domain and range of real functions
https://youtu.be/82LJ5MXAKRQ
5- Identical functions
https://youtu.be/ZOIt5JxoBxo
The reference book for these slides is
A Transition to Advanced Mathematics 8th Edition,
by Douglas Smith, Maurice Eggen, Richard St. Andre. ISBN-13: 978-1285463261, published by Cengage Learning (August 6, 2014).
https://www.cengagebrain.co.uk/shop/i...
The document discusses the distance formula and how to calculate the distance between two points. It provides the formula: Distance = √(x2 - x1)2 + (y2 - y1)2. Several examples are shown of using the distance formula to find the distance between points. The document also covers finding the midpoint between two points using the formula: Midpoint = (x1 + x2)/2, (y1 + y2)/2.
The document discusses the distance formula for finding the distance between two points P and Q on a line. It states that the distance is equal to the absolute value of P - Q. It provides an example where P = -2 and Q = 3, and calculates the distance as 5 using the formula |P - Q|. It then explains that we can square the difference P - Q to avoid issues with the absolute value notation, so the distance formula is the square root of (P - Q)2.
The document explains how to calculate the distance between two points using the distance formula. It shows that if points are horizontal or vertical, you can use their x- or y-coordinates alone, but otherwise you need to use the Pythagorean theorem to form a right triangle and find the hypotenuse. The distance formula is given as the square root of the sum of the squared differences between corresponding x- and y-coordinates of the two points. An example using points (3, -5) and (-1, 4) demonstrates applying the formula to find the distance between two points.
These slides cover the relation between two sets, domain and range of a relation, the inverse of a relation, and composition of relations.
Videos explaining these slides are available
Part1: Definition of relations https://youtu.be/MR0ALeKvaSc
Part2: Domain and range of a relation https://youtu.be/kBlUWTDn-Z4
Part3: The inverse of a relation https://youtu.be/Uv5KFusvvsY
Part4: Composition of relations https://youtu.be/IvVIjJqqPH0
Reference for these slides are
A Transition to Advanced Mathematics 8th Edition,
by Douglas Smith, Maurice Eggen, Richard St. Andre. ISBN-13: 978-1285463261, published by Cengage Learning (August 6, 2014).
https://www.cengagebrain.co.uk/shop/i...
This document provides formulas and examples for calculating the distance and midpoint between two points.
The distance formula is used to find the distance between two points (x1, y1) and (x2, y2). The midpoint formula finds the midpoint between two points by taking the average of the x-coordinates and y-coordinates.
Several examples demonstrate using the distance and midpoint formulas to find distances, midpoints, and missing endpoint coordinates. The key steps are to work inside parentheses, do exponents before addition, and plug points into the formulas.
1. The document discusses the distance and midpoint formulas in coordinate geometry.
2. It provides the formulas for finding the distance between two points and the midpoint of a line segment.
3. Examples are given of using the formulas to find distances and midpoints, as well as classifying triangles and finding endpoint coordinates given a midpoint.
1. The document discusses finding the midpoint and distance between points in a coordinate plane. It provides examples and steps for using the midpoint formula and distance formula.
2. The midpoint formula is used to find the midpoint between two points by taking the average of the x-coordinates and y-coordinates. The distance formula and Pythagorean theorem can both be used to find the distance between points.
3. Examples include finding midpoints and distances between points, as well as applications to sports such as finding the distance of a throw between bases in a baseball diamond.
1) The document discusses distance and midpoints between two points in coordinate geometry. It defines the distance formula as the absolute value of x2 - x1 plus y2 - y1.
2) It presents the midpoint formulas, where the x-coordinate of the midpoint is (x1 + x2)/2 and the y-coordinate is (y1 + y2)/2.
3) Examples are given to find the distance between two points, and to determine the midpoint or endpoints of a line segment given coordinates of two points.
This document introduces the distance formula, which is used to calculate the distance between two points (x1, y1) and (x2, y2) on a coordinate plane. The distance formula is the square root of (x1 - x2) squared plus (y1 - y2) squared. Several examples are worked through to demonstrate finding the distance between points using their coordinates. Practice problems are also provided for the reader to work through on their own.
The document discusses two methods for simplifying complex fractions. A complex fraction is a fraction with fractions in the numerator or denominator. The first method reduces the complex fraction to an "easy" problem by combining the numerator and denominator terms into single fractions. The second method multiplies the lowest common denominator of all terms to both the numerator and denominator of the complex fraction. An example using each method is provided to demonstrate the simplification process.
This an animated slides for students. Introduce basis concept of proofs to students. Direct proofs. Please search for slides Proof methods-teachers. If you want to teach using these slides.
The document discusses using the distance formula to calculate the distance between two points on a coordinate plane. It gives the formula for distance as the square root of (x1 - x2) squared plus (y1 - y2) squared, where (x1, y1) and (x2, y2) are the coordinates of the two points. It provides an example of using the distance formula and exercises for the reader to practice calculating distances between points.
The document discusses multiplying and dividing variable expressions. It provides examples of simplifying expressions using the distributive property and the property of the opposite of a sum. It also demonstrates dividing variable expressions by writing the division as a fraction and simplifying. Key steps include distributing terms, dividing each term in the numerator by the denominator, and evaluating expressions for given variable values.
11 x1 t11 06 tangents & normals ii (2013)Nigel Simmons
This document discusses finding tangents and normals to parabolas defined by equations of the form y = ax^2 using Cartesian coordinates. It shows that the slope of a tangent line at a point (x1, y1) is 2x1/a, and the slope of the corresponding normal is -a/x1. It also describes how to determine whether a given line will cut, touch, or miss the parabola based on the discriminant of the equation for their intersection.
The document discusses tests for determining if triangles are similar. There are three tests: (1) corresponding sides are in proportion (SSS), (2) two pairs of corresponding sides are in proportion and included angles are equal (SAS), (3) all three angles are equal (AA). An example problem finds the length of side AD using the properties of similar triangles. The side lengths are in the same ratio as the corresponding sides between the two triangles.
The document discusses finding the greatest term in polynomial expansions. It provides an example of finding the greatest coefficient in the expansion of (2 + 3x)^20. Through algebraic steps, it is shown that the greatest coefficient is T13 = 20C12 * 2831^2. It then gives an example of finding the greatest term in the expansion of (3x - 4)^15 when x = 1/2. After setting up expressions for the terms Tk+1 and Tk, it derives an inequality to determine the value of k that makes Tk+1 the greatest term.
The document discusses congruent triangles and the different tests that can be used to prove triangles are congruent. There are four main tests: side-side-side (SSS), side-angle-side (SAS), angle-angle-side (AAS), and right angle-hypotenuse-side (RHS). It also provides examples of applying the SAS test to prove corresponding sides and angles of two triangles are congruent. Additionally, it defines different types of triangles such as isosceles and equilateral triangles based on their side lengths and angle measures.
11 x1 t07 04 quadrilateral family (2012)Nigel Simmons
The document presents a diagram showing the hierarchical classification of quadrilaterals. It begins with quadrilateral as the parent shape and branches down to more specific shapes such as squares, rectangles, rhombi, parallelograms, trapezoids, and kites. The text explains that quadrilaterals can be classified based on their properties, and moving down the hierarchy, the shapes become more specialized. It also notes that any property of the parent shapes also applies to the child shapes.
The document discusses properties of transversals and ratios of intercepts formed when a transversal cuts parallel lines. It includes examples of calculating lengths of intercepts using given ratios and applying the ratio property that the ratio of corresponding intercepts is equal to the ratio of distances from the transversal to the parallel lines.
11 x1 t05 06 line through pt of intersection (2013)Nigel Simmons
The document provides steps to find the equation of a line passing through the intersection of two other lines and a given point. It first finds the intersection point of the two initial lines as (-2,3). It then calculates the slope of the line between this point and the given point (1,2) as -3. Finally, it derives the equation of the line as 7x + 21y - 49 = 0.
The document discusses key concepts in Euclidean geometry including definitions, notation, terminology, naming conventions, angle theorems, and constructing proofs. It defines parallel lines, perpendicular lines, congruent lines, and similar lines using notation symbols. It provides examples of naming angles, polygons, and parallel lines in a consistent cyclic order. It also outlines the steps and logical structure required to properly construct geometric proofs.
The document discusses the slope (gradient) of a line and how to calculate it. It provides four methods to calculate slope:
(1) The rise over the run between two points (vertical change over horizontal change)
(2) The change in y-values over the change in x-values between two points using a formula
(3) The slope of a line is equal to the tangent of the angle of inclination
(4) The relationship between slopes of parallel and perpendicular lines. Two lines are parallel if their slopes are equal, and perpendicular if the product of their slopes is -1. An example problem demonstrates finding the value of a that results in two lines being parallel or perpendicular.
The document discusses several triangle and polygon theorems:
- The sum of the interior angles of any triangle is 180 degrees.
- The exterior angle of a triangle is equal to the sum of the two opposite interior angles.
- The sum of the interior angles of any quadrilateral is 360 degrees.
- The sum of the interior angles of any pentagon is 540 degrees.
Proofs of these theorems are presented using angle properties, parallel lines, and adding individual triangle angle sums.
11 x1 t05 04 point slope formula (2013)Nigel Simmons
The document discusses the point slope formula and provides examples of its use. Specifically, it contains:
1) The point slope formula: y - y1 = m(x - x1)
2) An example that finds the equation of the line passing through (-3,4) and (2,-6).
3) A second example that finds the equation (3x + 4y + 6 = 0) of the line passing through (2,-3) and parallel to the given line (3x + 4y - 5 = 0).
All straight lines can be written in the form of y = mx + b, where m is the slope and b is the y-intercept. Alternatively, lines can be written in the general form of Ax + By + C = 0, where A, B, and C are integers or surds. Lines parallel to the x-axis have the form y = c, where c is a constant. Lines parallel to the y-axis have the form x = k, where k is a constant. The example shows finding the equation of a line perpendicular to another line in general form.
The document discusses several key concepts about triangles:
- The sum of the angles of any triangle is always 180 degrees.
- It defines different types of triangles based on angle measures, such as acute, obtuse, right, isosceles, equilateral, and scalene triangles.
- Parallel lines and auxiliary lines are used to help prove theorems about triangles. The exterior angle theorem states that the measure of an exterior angle is equal to the sum of the remote interior angles.
11 x1 t08 03 angle between two lines (2013)Nigel Simmons
This document describes how to find the acute angle between two lines given their slopes m1 and m2. The method is to take the tangent of both sides of the equation m1/
11X1 T05 01 division of an interval (2011)Nigel Simmons
Here are the steps to solve this problem:
1) Write down the endpoints of the interval: (-3,4) and (5,6)
2) The ratio given is 1:3
3) Let the interval from the first endpoint to P be 1 unit
4) Then the interval from P to the second endpoint is 3 units
5) Use the ratio of distances formula to find the coordinates of P
1) The point midway between -7 and 5 is -1.
2) The point midway between (-3, 1) and (5, 4) is (1, 2.5).
3) The distance between points (1, 1) and (4, 5) is 5.
The student is able to (I can):
• Find the midpoint of two given points.
• Find the coordinates of an endpoint given one endpoint
and a midpoint.
• Find the distance between two points.
1. The document discusses distance and midpoints between two points. It provides the distance formula and explains how to find the distance and midpoint between two points given their x and y coordinates.
2. It gives examples of using the distance formula to calculate the distance between points like (-3,2) and (1,-1), which is √25 or 5 units.
3. The midpoint between two points is calculated by taking the average of the x-coordinates and the average of the y-coordinates. For points (7,2) and (-3,6), the midpoint is (2,4).
This document discusses formulas for distance, midpoint, and the Pythagorean theorem. It provides the formulas to find the distance between two points, the midpoint of a line segment, and the length of the hypotenuse of a right triangle. It includes examples of using the formulas to find distances, midpoints, and missing lengths in right triangles. It assigns homework problems involving using these formulas to solve geometric problems.
The document discusses distance and midpoint formulas for calculating the distance between two points and finding the midpoint of two points. It provides examples of using the distance formula, d = √(x2 - x1)2 + (y2 - y1)2, and midpoint formula, (x1 + x2)/2, (y1 + y2)/2. It also covers using these formulas to find distances, midpoints, and unknown endpoints of line segments between points.
This document provides information on calculating distance between two points using the distance formula. It gives the distance formula, (x1 - x2)2 + (y1 - y2)2, and provides an example of using it to find the distance between points F(3, 2) and G(-3, -1), which equals 6.7. It also gives two practice problems and their solutions: 1) the distance between (9, -1) and (6, 3) is 5, and 2) the distance between points R(10, 15) and S(6, 20) is 41.
(8) Lesson 5.7 - Distance on the Coordinate Planewzuri
This document provides examples for applying algebraic concepts to geometry problems. It demonstrates using the Pythagorean theorem and distance formula to calculate distances between points on a coordinate plane. Sample problems are worked through step-by-step and answers are provided. Common Core standards addressed include applying the Pythagorean theorem to find distances between points and using square root and cube root symbols to represent solutions to equations.
The document discusses finding the midpoint and distance between two points with given coordinates. It provides formulas for finding the midpoint, which is the average of the x-coordinates and y-coordinates, and the distance, which uses the difference of the x-coordinates and y-coordinates. Several examples demonstrate using these formulas to calculate midpoints and distances. Practice problems with solutions are also provided.
1. The document discusses functions and sequences. It begins by defining a function as an assignment of exactly one element from the codomain to each element of the domain.
2. It provides examples of determining the domain, codomain, and range of various functions. It also discusses one-to-one functions, where each element of the domain is mapped to a unique element in the codomain.
3. The document discusses adding and multiplying functions by defining (f1 + f2)(x) = f1(x) + f2(x) and (f1f2)(x) = f1(x)f2(x). It provides examples of calculating the sum and product of various
The document summarizes the Pythagorean theorem and distance formula. It states that the Pythagorean theorem equation is a^2 + b^2 = c^2, where a and b are the legs of a right triangle and c is the hypotenuse. An example problem demonstrates solving for the missing side of a right triangle. The distance formula is defined as the square root of (x2-x1)^2 + (y2-y1)^2, and a worked example finds the distance between two points.
The document discusses the Pythagorean theorem and distance formula. It defines the Pythagorean theorem equation as a^2 + b^2 = c^2, where a and b are the legs of a right triangle and c is the hypotenuse. An example problem demonstrates using the theorem to solve for the missing side of a right triangle. The distance formula is defined as the square root of (x2-x1)^2 + (y2-y1)^2, and an example shows labeling points and plugging them into the formula to calculate the distance between two points.
The document describes how to solve simultaneous equations using three steps: 1) eliminate a variable from the equations, 2) solve for the remaining variable, and 3) substitute back to find the eliminated variable. It provides an example showing these steps to solve two simultaneous equations for x and y. It then shows another example involving eliminating variables to create two new equations that can be solved simultaneously for y and z. The key points are eliminating variables to reduce the equations, then solving the reduced equations to find the values of the variables.
The document describes how to solve simultaneous equations using three steps: 1) eliminate a variable from the equations, 2) solve for the remaining variable, and 3) substitute back to find the eliminated variable. It provides an example showing these steps to solve two simultaneous equations for x and y. It then shows another example involving eliminating variables to create two new equations that can be solved simultaneously for y and z. The key point is that simultaneous equations can be solved when there are as many equations as unknown variables.
The document describes how to solve simultaneous equations using three steps: 1) eliminate a variable from the equations, 2) solve for the remaining variable, and 3) substitute back to find the eliminated variable. It provides an example problem demonstrating these steps, showing the elimination of variables through multiplication and addition of the equations until a single variable remains that can be solved for. The document notes that simultaneous equations with the same number of equations as variables can always be solved to find the values of the variables.
The document describes how to solve simultaneous equations using three steps: 1) eliminate a variable from the equations, 2) solve for the remaining variable, and 3) substitute back to find the eliminated variable. It provides an example problem demonstrating these steps, eliminating variables through multiplication and addition of the equations until a single variable remains that can be solved for. The document notes that simultaneous equations with the same number of variables and equations can always be solved using this method.
Here are the steps to solve this revision exercise:
1. Use the distance formula to find the lengths of each side:
a) AB = √(−1 − 2)2 + (4 − 3)2 = 3.16
b) BC = √(1 − (−1))2 + (−3 − 4)2 = 7.28
c) AC = √(1 − 2)2 + (−3 − 3)2 = 6.08
2. Use the midpoint formula to find the midpoints of each side:
a) MPAB = (0.5, 3.5)
b) MPBC = (0, 0.5)
c) MPAC = (
The document discusses the Pythagorean theorem and distance formula. The Pythagorean theorem states that for any right triangle, the sum of the squares of the two sides equals the square of the hypotenuse. The distance formula allows you to calculate the distance between two points in the xy-plane by taking the square root of the sum of the squared differences between the x-coordinates and y-coordinates. An example using points (2,4) and (7,13) demonstrates applying the distance formula to get a distance of 10.3 units between the two points.
The document provides information about finding the coordinates of points, plotting pairs of points, calculating the slope of lines between points, and obtaining the equation of a line using the two-point form. It gives examples of finding the slope and equation of lines passing through various pairs of points. It also includes examples for students to practice finding the equations of lines from given points.
Similar to 11 x1 t05 01 division of an interval (2013) (20)
Slideshare is discontinuing its Slidecast feature as of February 28, 2014. Existing Slidecasts will be converted to static presentations without audio by April 30, 2014. The document informs users that new slidecasts can be found on myPlick.com or the author's blog starting in 2014. However, myPlick proved unreliable, so future slidecasts will instead be hosted on the author's YouTube channel.
The document discusses different methods for factorising expressions:
1) Looking for a common factor and dividing it out of all terms
2) Using the difference of two squares formula (a2 - b2 = (a - b)(a + b))
3) Factorising quadratic trinomials into two binomial factors by identifying the values that multiply to give the constant term and sum to give the coefficient of the linear term.
The document provides information on index laws and the meaning of indices in algebra:
- Index laws state that am × an = am+n, am ÷ an = am-n, and (am)n = amn. Exponents can be added or subtracted when multiplying or dividing terms with the same base.
- Positive exponents indicate a term is raised to a power. Negative exponents indicate a root is being taken. Terms with exponents are evaluated from left to right.
- Examples demonstrate how to simplify expressions using index laws and interpret different types of indices.
12 x1 t01 03 integrating derivative on function (2013)Nigel Simmons
The document discusses integrating derivatives of functions. It states that the integral of the derivative of a function f(x) is equal to the natural log of f(x) plus a constant. It then provides examples of integrating several derivatives: (i) ∫(1/(7-3x)) dx = -1/3 log(7-3x) + c, (ii) ∫(1/(8x+5)) dx = 1/8 log(8x+5) + c, and (iii) ∫(x5/(x-2)) dx = 1/6 log(x6-2) + c. It also discusses techniques for integrating fractions by polynomial long division and finds
The document discusses logarithms and their properties. Logarithms are defined as the inverse of exponentials. If y = ax, then x = loga y. The natural logarithm is log base e, written as ln. Properties of logarithms include: loga m + loga n = loga mn; loga m - loga n = loga(m/n); loga mn = n loga m; loga 1 = 0; loga a = 1. Examples of evaluating logarithmic expressions are provided.
The document discusses relationships between the coefficients and roots of polynomials. It states that for a polynomial P(x) = axn + bxn-1 + cxn-2 + ..., the sum of the roots equals -b/a, the sum of the roots taken two at a time equals c/a, and so on for higher order terms. It also provides examples of using these relationships to find the sums of roots for a given polynomial.
P
4
3
2
The document discusses properties of polynomials with multiple roots. It first proves that if a polynomial P(x) has a root x = a of multiplicity m, then the derivative of P(x), P'(x), will have a root x = a of multiplicity m-1. It then provides an example of solving a cubic equation given it has a double root. Finally, it examines a quartic polynomial and shows that its root α cannot be 0, 1, or -1, and that 1/
The document discusses factorizing complex expressions. The main points are:
- If a polynomial's coefficients are real, its roots will appear in complex conjugate pairs.
- Any polynomial of degree n can be factorized into a mixture of quadratic and linear factors over real numbers, or into n linear factors over complex numbers.
- Odd degree polynomials must have at least one real root.
- Examples of factorizing polynomials over both real and complex numbers are provided.
The document describes the Trapezoidal Rule for approximating the area under a curve between two points. It shows that the area A is estimated by dividing the region into trapezoids with height equal to the function values at the interval endpoints and bases equal to the intervals. In general, the area is approximated as the sum of the areas of each trapezoid, which is equal to the average of the endpoint function values multiplied by the interval length.
The document discusses methods for calculating the volumes of solids of revolution. It provides formulas for finding volumes when an area is revolved around either the x-axis or y-axis. Examples are given for finding volumes of common solids like cones, spheres, and others. Steps are shown for using the formulas to calculate volumes based on given functions and limits of revolution.
The document discusses different methods for calculating the area under a curve or between curves.
(1) The area below the x-axis is given by the integral of the function between the bounds, which can be positive or negative depending on whether the area is above or below the x-axis.
(2) To calculate the area on the y-axis, the function is solved for x in terms of y, then the bounds are substituted into the integral of this new function with respect to y.
(3) The area between two curves is calculated by taking the integral of the upper curve minus the integral of the lower curve, both between the same bounds on the x-axis.
The document discusses 8 properties of definite integrals:
1) Integrating polynomials results in a fraction.
2) Constants can be factored out of integrals.
3) Integrals of sums are equal to the sum of integrals.
4) Splitting an integral range results in the sum of the integrals.
5) Integrals of positive functions over a range are positive, and negative if the function is negative.
6) Integrals can be compared based on the relative values of the integrands.
7) Changing the limits of integration flips the sign of the integral.
8) Integrals of odd functions over a symmetric range are zero, and integrals of even functions are twice the integral over
Elevate Your Nonprofit's Online Presence_ A Guide to Effective SEO Strategies...TechSoup
Whether you're new to SEO or looking to refine your existing strategies, this webinar will provide you with actionable insights and practical tips to elevate your nonprofit's online presence.
Andreas Schleicher presents PISA 2022 Volume III - Creative Thinking - 18 Jun...EduSkills OECD
Andreas Schleicher, Director of Education and Skills at the OECD presents at the launch of PISA 2022 Volume III - Creative Minds, Creative Schools on 18 June 2024.
This presentation was provided by Racquel Jemison, Ph.D., Christina MacLaughlin, Ph.D., and Paulomi Majumder. Ph.D., all of the American Chemical Society, for the second session of NISO's 2024 Training Series "DEIA in the Scholarly Landscape." Session Two: 'Expanding Pathways to Publishing Careers,' was held June 13, 2024.
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إضغ بين إيديكم من أقوى الملازم التي صممتها
ملزمة تشريح الجهاز الهيكلي (نظري 3)
💀💀💀💀💀💀💀💀💀💀
تتميز هذهِ الملزمة بعِدة مُميزات :
1- مُترجمة ترجمة تُناسب جميع المستويات
2- تحتوي على 78 رسم توضيحي لكل كلمة موجودة بالملزمة (لكل كلمة !!!!)
#فهم_ماكو_درخ
3- دقة الكتابة والصور عالية جداً جداً جداً
4- هُنالك بعض المعلومات تم توضيحها بشكل تفصيلي جداً (تُعتبر لدى الطالب أو الطالبة بإنها معلومات مُبهمة ومع ذلك تم توضيح هذهِ المعلومات المُبهمة بشكل تفصيلي جداً
5- الملزمة تشرح نفسها ب نفسها بس تكلك تعال اقراني
6- تحتوي الملزمة في اول سلايد على خارطة تتضمن جميع تفرُعات معلومات الجهاز الهيكلي المذكورة في هذهِ الملزمة
واخيراً هذهِ الملزمة حلالٌ عليكم وإتمنى منكم إن تدعولي بالخير والصحة والعافية فقط
كل التوفيق زملائي وزميلاتي ، زميلكم محمد الذهبي 💊💊
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A Visual Guide to 1 Samuel | A Tale of Two HeartsSteve Thomason
These slides walk through the story of 1 Samuel. Samuel is the last judge of Israel. The people reject God and want a king. Saul is anointed as the first king, but he is not a good king. David, the shepherd boy is anointed and Saul is envious of him. David shows honor while Saul continues to self destruct.
Level 3 NCEA - NZ: A Nation In the Making 1872 - 1900 SML.pptHenry Hollis
The History of NZ 1870-1900.
Making of a Nation.
From the NZ Wars to Liberals,
Richard Seddon, George Grey,
Social Laboratory, New Zealand,
Confiscations, Kotahitanga, Kingitanga, Parliament, Suffrage, Repudiation, Economic Change, Agriculture, Gold Mining, Timber, Flax, Sheep, Dairying,
6. Coordinate Geometry
Distance Formula
d x2 x1 y2 y1
2 2
e.g. Find the distance between
(–1,3) and (3,5)
d 5 3 3 1
2 2
22 42
20
2 5 units
7. Coordinate Geometry
Distance Formula
d x2 x1 y2 y1
2 2
e.g. Find the distance between
(–1,3) and (3,5)
d 5 3 3 1
2 2
22 42
20
2 5 units
The distance formula is
finding the length of the hypotenuse,
using Pythagoras
8. Coordinate Geometry
Distance Formula Midpoint Formula
d x2 x1 y2 y1
2 2
e.g. Find the distance between
(–1,3) and (3,5)
d 5 3 3 1
2 2
22 42
20
2 5 units
The distance formula is
finding the length of the hypotenuse,
using Pythagoras
9. Coordinate Geometry
Distance Formula Midpoint Formula
x1 x2 y1 y2
d x2 x1 y2 y1
2 2
M ,
2 2
e.g. Find the distance between
(–1,3) and (3,5)
d 5 3 3 1
2 2
22 42
20
2 5 units
The distance formula is
finding the length of the hypotenuse,
using Pythagoras
10. Coordinate Geometry
Distance Formula Midpoint Formula
x1 x2 y1 y2
d x2 x1 y2 y1
2 2
M ,
2 2
e.g. Find the distance between e.g. Find the midpoint of
(–1,3) and (3,5) (3,4) and (–2 ,1)
d 5 3 3 1
2 2
22 42
20
2 5 units
The distance formula is
finding the length of the hypotenuse,
using Pythagoras
11. Coordinate Geometry
Distance Formula Midpoint Formula
x1 x2 y1 y2
d x2 x1 y2 y1
2 2
M ,
2 2
e.g. Find the distance between e.g. Find the midpoint of
(–1,3) and (3,5) (3,4) and (–2 ,1)
d 5 3 3 1
2 2
3 2 , 4 1
M
2 2
22 42
20
2 5 units
The distance formula is
finding the length of the hypotenuse,
using Pythagoras
12. Coordinate Geometry
Distance Formula Midpoint Formula
x1 x2 y1 y2
d x2 x1 y2 y1
2 2
M ,
2 2
e.g. Find the distance between e.g. Find the midpoint of
(–1,3) and (3,5) (3,4) and (–2 ,1)
d 5 3 3 1
2 2
3 2 , 4 1
M
2 2
22 42
,
1 5
20
2 2
2 5 units
The distance formula is
finding the length of the hypotenuse,
using Pythagoras
13. Coordinate Geometry
Distance Formula Midpoint Formula
x1 x2 y1 y2
d x2 x1 y2 y1
2 2
M ,
2 2
e.g. Find the distance between e.g. Find the midpoint of
(–1,3) and (3,5) (3,4) and (–2 ,1)
d 5 3 3 1
2 2
3 2 , 4 1
M
2 2
22 42
,
1 5
20
2 2
2 5 units
The midpoint formula
The distance formula is
averages the x and y values
finding the length of the hypotenuse,
using Pythagoras
15. Division Of An Interval
Mathematics (2 unit) division of an interval questions are restricted
to midpoint questions i.e. dividing in the ratio 1:1
16. Division Of An Interval
Mathematics (2 unit) division of an interval questions are restricted
to midpoint questions i.e. dividing in the ratio 1:1
In Extension 1 you can be asked to divide an interval in a any
ratio, and it could be either an internal or an external division.
17. Division Of An Interval
Mathematics (2 unit) division of an interval questions are restricted
to midpoint questions i.e. dividing in the ratio 1:1
In Extension 1 you can be asked to divide an interval in a any
ratio, and it could be either an internal or an external division.
X Y
18. Division Of An Interval
Mathematics (2 unit) division of an interval questions are restricted
to midpoint questions i.e. dividing in the ratio 1:1
In Extension 1 you can be asked to divide an interval in a any
ratio, and it could be either an internal or an external division.
X A Y
19. Division Of An Interval
Mathematics (2 unit) division of an interval questions are restricted
to midpoint questions i.e. dividing in the ratio 1:1
In Extension 1 you can be asked to divide an interval in a any
ratio, and it could be either an internal or an external division.
A divides XY internally
in the ratio m:n
X A Y
20. Division Of An Interval
Mathematics (2 unit) division of an interval questions are restricted
to midpoint questions i.e. dividing in the ratio 1:1
In Extension 1 you can be asked to divide an interval in a any
ratio, and it could be either an internal or an external division.
A divides XY internally
in the ratio m:n
X A Y
m n
21. Division Of An Interval
Mathematics (2 unit) division of an interval questions are restricted
to midpoint questions i.e. dividing in the ratio 1:1
In Extension 1 you can be asked to divide an interval in a any
ratio, and it could be either an internal or an external division.
A divides XY internally
in the ratio m:n
X A Y OR
m n
22. Division Of An Interval
Mathematics (2 unit) division of an interval questions are restricted
to midpoint questions i.e. dividing in the ratio 1:1
In Extension 1 you can be asked to divide an interval in a any
ratio, and it could be either an internal or an external division.
A divides XY internally
in the ratio m:n
X A Y OR
A divides YX internally
m n
in the ratio n:m
23. Division Of An Interval
Mathematics (2 unit) division of an interval questions are restricted
to midpoint questions i.e. dividing in the ratio 1:1
In Extension 1 you can be asked to divide an interval in a any
ratio, and it could be either an internal or an external division.
A divides XY internally
in the ratio m:n
X A Y OR
A divides YX internally
m n
in the ratio n:m
X Y
24. Division Of An Interval
Mathematics (2 unit) division of an interval questions are restricted
to midpoint questions i.e. dividing in the ratio 1:1
In Extension 1 you can be asked to divide an interval in a any
ratio, and it could be either an internal or an external division.
A divides XY internally
in the ratio m:n
X A Y OR
A divides YX internally
m n
in the ratio n:m
X Y A
25. Division Of An Interval
Mathematics (2 unit) division of an interval questions are restricted
to midpoint questions i.e. dividing in the ratio 1:1
In Extension 1 you can be asked to divide an interval in a any
ratio, and it could be either an internal or an external division.
A divides XY internally
in the ratio m:n
X A Y OR
A divides YX internally
m n
in the ratio n:m
A divides XY externally
X Y A in the ratio m:n
26. Division Of An Interval
Mathematics (2 unit) division of an interval questions are restricted
to midpoint questions i.e. dividing in the ratio 1:1
In Extension 1 you can be asked to divide an interval in a any
ratio, and it could be either an internal or an external division.
A divides XY internally
in the ratio m:n
X A Y OR
A divides YX internally
m n
in the ratio n:m
n
A divides XY externally
X Y A in the ratio m:n
m
27. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
28. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
29. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
3,4 5,6
30. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
• Write down the ratio.
3,4 5,6
31. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
• Write down the ratio.
3,4 5,6
1: 3
32. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
• Write down the ratio.
• Draw a cross joining the ratio to the two points
3,4 5,6
1: 3
33. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
• Write down the ratio.
• Draw a cross joining the ratio to the two points
3,4 5,6
1: 3
34. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
• Write down the ratio.
• Draw a cross joining the ratio to the two points
• Set up your answer by drawing a set of parentheses with two
vinculums separated by a comma
3,4 5,6
1: 3
35. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
• Write down the ratio.
• Draw a cross joining the ratio to the two points
• Set up your answer by drawing a set of parentheses with two
vinculums separated by a comma
3,4 5,6
P ,
1: 3
36. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
• Write down the ratio.
• Draw a cross joining the ratio to the two points
• Set up your answer by drawing a set of parentheses with two
vinculums separated by a comma
• Add the numbers in the ratio together to get the denominator
3,4 5,6
P ,
1: 3
37. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
• Write down the ratio.
• Draw a cross joining the ratio to the two points
• Set up your answer by drawing a set of parentheses with two
vinculums separated by a comma
• Add the numbers in the ratio together to get the denominator
3,4 5,6
P ,
4 4
1: 3
38. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
• Write down the ratio.
• Draw a cross joining the ratio to the two points
• Set up your answer by drawing a set of parentheses with two
vinculums separated by a comma
• Add the numbers in the ratio together to get the denominator
• Multiply along the cross and add to get the numerator
3,4 5,6 P ,
4 4
1: 3
39. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
• Write down the ratio.
• Draw a cross joining the ratio to the two points
• Set up your answer by drawing a set of parentheses with two
vinculums separated by a comma
• Add the numbers in the ratio together to get the denominator
• Multiply along the cross and add to get the numerator
3,4 5,6 P 3 3 1 5 ,
4 4
1: 3
40. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
• Write down the ratio.
• Draw a cross joining the ratio to the two points
• Set up your answer by drawing a set of parentheses with two
vinculums separated by a comma
• Add the numbers in the ratio together to get the denominator
• Multiply along the cross and add to get the numerator
3,4 5,6 P 3 3 1 5 , 3 4 1 6
4 4
1: 3
41. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
• Write down the ratio.
• Draw a cross joining the ratio to the two points
• Set up your answer by drawing a set of parentheses with two
vinculums separated by a comma
• Add the numbers in the ratio together to get the denominator
• Multiply along the cross and add to get the numerator
3,4 5,6 P 3 3 1 5 , 3 4 1 6
4 4
4 , 18
1: 3
4 4
42. Type 1: Internal Division 2003 Extension 1 HSC Q1c)
Find the coordinates of P that divides the interval joining 3,4 and
5,6 internally in the ratio 1 : 3
• Write down the endpoints of the interval in the same order as
they are mentioned.
• Write down the ratio.
• Draw a cross joining the ratio to the two points
• Set up your answer by drawing a set of parentheses with two
vinculums separated by a comma
• Add the numbers in the ratio together to get the denominator
• Multiply along the cross and add to get the numerator
3,4 5,6 P 3 3 1 5 , 3 4 1 6
4 4
4 , 18
1, 9
1: 3
4 4 2
43. Type 2: External Division 2004 Extension 1 HSC Q1c)
Let A be the point 3,1 and B be the point 9,2 . Find the coordinates
of the point P which divides AB externally in the ratio 5 : 2.
44. Type 2: External Division 2004 Extension 1 HSC Q1c)
Let A be the point 3,1 and B be the point 9,2 . Find the coordinates
of the point P which divides AB externally in the ratio 5 : 2.
• Done exactly the same as internal division, except make one of
the numbers in the ratio negative
45. Type 2: External Division 2004 Extension 1 HSC Q1c)
Let A be the point 3,1 and B be the point 9,2 . Find the coordinates
of the point P which divides AB externally in the ratio 5 : 2.
• Done exactly the same as internal division, except make one of
the numbers in the ratio negative
3,1 9,2
46. Type 2: External Division 2004 Extension 1 HSC Q1c)
Let A be the point 3,1 and B be the point 9,2 . Find the coordinates
of the point P which divides AB externally in the ratio 5 : 2.
• Done exactly the same as internal division, except make one of
the numbers in the ratio negative
3,1 9,2
5:2
47. Type 2: External Division 2004 Extension 1 HSC Q1c)
Let A be the point 3,1 and B be the point 9,2 . Find the coordinates
of the point P which divides AB externally in the ratio 5 : 2.
• Done exactly the same as internal division, except make one of
the numbers in the ratio negative
3,1 9,2
5:2
48. Type 2: External Division 2004 Extension 1 HSC Q1c)
Let A be the point 3,1 and B be the point 9,2 . Find the coordinates
of the point P which divides AB externally in the ratio 5 : 2.
• Done exactly the same as internal division, except make one of
the numbers in the ratio negative
3,1 9,2
5:2
P ,
49. Type 2: External Division 2004 Extension 1 HSC Q1c)
Let A be the point 3,1 and B be the point 9,2 . Find the coordinates
of the point P which divides AB externally in the ratio 5 : 2.
• Done exactly the same as internal division, except make one of
the numbers in the ratio negative
3,1 9,2
5:2
P ,
3 3
50. Type 2: External Division 2004 Extension 1 HSC Q1c)
Let A be the point 3,1 and B be the point 9,2 . Find the coordinates
of the point P which divides AB externally in the ratio 5 : 2.
• Done exactly the same as internal division, except make one of
the numbers in the ratio negative
3,1 9,2
5:2
2 3 5 9
P ,
3 3
51. Type 2: External Division 2004 Extension 1 HSC Q1c)
Let A be the point 3,1 and B be the point 9,2 . Find the coordinates
of the point P which divides AB externally in the ratio 5 : 2.
• Done exactly the same as internal division, except make one of
the numbers in the ratio negative
3,1 9,2
5:2
2 3 5 9 2 1 5 2
P ,
3 3
52. Type 2: External Division 2004 Extension 1 HSC Q1c)
Let A be the point 3,1 and B be the point 9,2 . Find the coordinates
of the point P which divides AB externally in the ratio 5 : 2.
• Done exactly the same as internal division, except make one of
the numbers in the ratio negative
3,1 9,2
5:2
2 3 5 9 2 1 5 2
P ,
3 3
39 , 12
3 3
53. Type 2: External Division 2004 Extension 1 HSC Q1c)
Let A be the point 3,1 and B be the point 9,2 . Find the coordinates
of the point P which divides AB externally in the ratio 5 : 2.
• Done exactly the same as internal division, except make one of
the numbers in the ratio negative
3,1 9,2
5:2
2 3 5 9 2 1 5 2
P ,
3 3
39 , 12
3 3
13,4
54. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
55. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
• Draw the endpoints, ratio and cross the same as previously
56. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
• Draw the endpoints, ratio and cross the same as previously
1,8 x, y
57. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
• Draw the endpoints, ratio and cross the same as previously
1,8 x, y
2:3
58. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
• Draw the endpoints, ratio and cross the same as previously
1,8 x, y
2:3
59. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
• Draw the endpoints, ratio and cross the same as previously
• Create the fraction for the x value and equate it with the known value
1,8 x, y
2:3
60. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
• Draw the endpoints, ratio and cross the same as previously
• Create the fraction for the x value and equate it with the known value
1,8 x, y
3 1 2 x 2:3
1
5
61. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
• Draw the endpoints, ratio and cross the same as previously
• Create the fraction for the x value and equate it with the known value
1,8 x, y
3 1 2 x 2:3
1
5
5 3 2 x
62. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
• Draw the endpoints, ratio and cross the same as previously
• Create the fraction for the x value and equate it with the known value
1,8 x, y
3 1 2 x 2:3
1
5
5 3 2 x
2x 8
x4
63. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
• Draw the endpoints, ratio and cross the same as previously
• Create the fraction for the x value and equate it with the known value
• Repeat for the y value
1,8 x, y
3 1 2 x 2:3
1
5
5 3 2 x
2x 8
x4
64. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
• Draw the endpoints, ratio and cross the same as previously
• Create the fraction for the x value and equate it with the known value
• Repeat for the y value
1,8 x, y
3 1 2 x 2:3 3 8 2 y
1 4
5 5
5 3 2 x
2x 8
x4
65. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
• Draw the endpoints, ratio and cross the same as previously
• Create the fraction for the x value and equate it with the known value
• Repeat for the y value
1,8 x, y
3 1 2 x 2:3 3 8 2 y
1 4
5 5
5 3 2 x 20 24 2 y
2x 8
x4
66. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
• Draw the endpoints, ratio and cross the same as previously
• Create the fraction for the x value and equate it with the known value
• Repeat for the y value
1,8 x, y
3 1 2 x 2:3 3 8 2 y
1 4
5 5
5 3 2 x 20 24 2 y
2x 8 2 y 4
x4 y 2
67. Type 3: Find an endpoint of an interval 2005 Extension 1 HSC Q1e)
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
• Draw the endpoints, ratio and cross the same as previously
• Create the fraction for the x value and equate it with the known value
• Repeat for the y value
1,8 x, y
3 1 2 x 2:3 3 8 2 y
1 4
5 5
5 3 2 x 20 24 2 y
2x 8 2 y 4
x4 B 4,2 y 2
68. Alternative
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
69. Alternative
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
A B
70. Alternative
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
If P divides AB internally in the ratio 2 : 3
A P B
2 3
71. Alternative
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
If P divides AB internally in the ratio 2 : 3
A P B Then B divides AP externally in the ratio
2 3 5:3
72. Alternative
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
If P divides AB internally in the ratio 2 : 3
A P B Then B divides AP externally in the ratio
2 3 5:3
1,8 1,4
73. Alternative
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
If P divides AB internally in the ratio 2 : 3
A P B Then B divides AP externally in the ratio
2 3 5:3
1,8 1,4
5:3
74. Alternative
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
If P divides AB internally in the ratio 2 : 3
A P B Then B divides AP externally in the ratio
2 3 5:3
1,8 1,4
5:3
75. Alternative
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
If P divides AB internally in the ratio 2 : 3
A P B Then B divides AP externally in the ratio
2 3 5:3
1,8 1,4 B ,
5:3
76. Alternative
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
If P divides AB internally in the ratio 2 : 3
A P B Then B divides AP externally in the ratio
2 3 5:3
1,8 1,4 B ,
2 2
5:3
77. Alternative
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
If P divides AB internally in the ratio 2 : 3
A P B Then B divides AP externally in the ratio
2 3 5:3
3 1 5 1
1,8 1,4 B ,
2 2
5:3
78. Alternative
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
If P divides AB internally in the ratio 2 : 3
A P B Then B divides AP externally in the ratio
2 3 5:3
3 1 5 1 3 8 5 4
1,8 1,4 B ,
2 2
5:3
79. Alternative
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
If P divides AB internally in the ratio 2 : 3
A P B Then B divides AP externally in the ratio
2 3 5:3
3 1 5 1 3 8 5 4
1,8 1,4 B ,
2 2
8, 4
5:3
2 2
80. Alternative
The point P1,4 divides the line segment joining A 1,8 and B x, y
internally in the ratio 2 : 3.
Find the coordinates of the point B.
If P divides AB internally in the ratio 2 : 3
A P B Then B divides AP externally in the ratio
2 3 5:3
3 1 5 1 3 8 5 4
1,8 1,4 B ,
2 2
8, 4
5:3
2 2
4,2
81. Type 4: Finding the ratio 1991 Extension 1 HSC Q1c)
The point P 3,8 divides the interval externally in the ratio k : 1.
If A is the point 6,4 and B is the point 0,4 find the value of k.
,
82. Type 4: Finding the ratio 1991 Extension 1 HSC Q1c)
The point P 3,8 divides the interval externally in the ratio k : 1.
If A is the point 6,4 and B is the point 0,4 find the value of k.
,
• Draw the endpoints, ratio and cross the same as usual
83. Type 4: Finding the ratio 1991 Extension 1 HSC Q1c)
The point P 3,8 divides the interval externally in the ratio k : 1.
If A is the point 6,4 and B is the point 0,4 find the value of k.
,
• Draw the endpoints, ratio and cross the same as usual
6,4 0,4
84. Type 4: Finding the ratio 1991 Extension 1 HSC Q1c)
The point P 3,8 divides the interval externally in the ratio k : 1.
If A is the point 6,4 and B is the point 0,4 find the value of k.
,
• Draw the endpoints, ratio and cross the same as usual
6,4 0,4
k :1
85. Type 4: Finding the ratio 1991 Extension 1 HSC Q1c)
The point P 3,8 divides the interval externally in the ratio k : 1.
If A is the point 6,4 and B is the point 0,4 find the value of k.
,
• Draw the endpoints, ratio and cross the same as usual
6,4 0,4
k :1
86. Type 4: Finding the ratio 1991 Extension 1 HSC Q1c)
The point P 3,8 divides the interval externally in the ratio k : 1.
If A is the point 6,4 and B is the point 0,4 find the value of k.
,
• Draw the endpoints, ratio and cross the same as usual
• Create the fraction for the either the x value or the y value (it does
not matter which one) and equate it with the known value
6,4 0,4
k :1
87. Type 4: Finding the ratio 1991 Extension 1 HSC Q1c)
The point P 3,8 divides the interval externally in the ratio k : 1.
If A is the point 6,4 and B is the point 0,4 find the value of k.
,
• Draw the endpoints, ratio and cross the same as usual
• Create the fraction for the either the x value or the y value (it does
not matter which one) and equate it with the known value
6,4 0,4
k :1
1 6 k 0
3
k 1
88. Type 4: Finding the ratio 1991 Extension 1 HSC Q1c)
The point P 3,8 divides the interval externally in the ratio k : 1.
If A is the point 6,4 and B is the point 0,4 find the value of k.
,
• Draw the endpoints, ratio and cross the same as usual
• Create the fraction for the either the x value or the y value (it does
not matter which one) and equate it with the known value
6,4 0,4
k :1
1 6 k 0
3
k 1
3k 3 6
89. Type 4: Finding the ratio 1991 Extension 1 HSC Q1c)
The point P 3,8 divides the interval externally in the ratio k : 1.
If A is the point 6,4 and B is the point 0,4 find the value of k.
,
• Draw the endpoints, ratio and cross the same as usual
• Create the fraction for the either the x value or the y value (it does
not matter which one) and equate it with the known value
6,4 0,4
k :1
1 6 k 0
3
k 1
3k 3 6
3k 9
k 3
90. Type 4: Finding the ratio 1991 Extension 1 HSC Q1c)
The point P 3,8 divides the interval externally in the ratio k : 1.
If A is the point 6,4 and B is the point 0,4 find the value of k.
,
• Draw the endpoints, ratio and cross the same as usual
• Create the fraction for the either the x value or the y value (it does
not matter which one) and equate it with the known value
6,4 0,4
k :1
1 6 k 0
3
k 1
3k 3 6
3k 9
k 3 P divides AB externally in the ratio 3 : 1
91. Type 4: Finding the ratio 1991 Extension 1 HSC Q1c)
The point P 3,8 divides the interval externally in the ratio k : 1.
If A is the point 6,4 and B is the point 0,4 find the value of k.
,
• Draw the endpoints, ratio and cross the same as usual
• Create the fraction for the either the x value or the y value (it does
not matter which one) and equate it with the known value
6,4 0,4
Exercise 5A;
1ad, 2ad,
k :1 3 i, iii in all, 4ace,
1 6 k 0
3 5 i, ii ace, 6bd,
k 1 8, 9, 11, 13b, 16,
3k 3 6 17, 20, 21, 23, 24
3k 9
k 3 P divides AB externally in the ratio 3 : 1