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Analytical geometry, basic concepts and calculations.

Analytical geometry, basic concepts and calculations.

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## Analytical geometry Grade 10Presentation Transcript

• ANALYTIC GEOMETRY GRADE: 10 Lajaé Plaatjies 201248629
• HISTORY Introduced in the 1630s , an important mathematical development, for it laid the foundations for modern mathematics as well as Aided the development of calculus RENE DESCARTES (1596-1650) and PIERRE DE FERMAT (1601-1665), French mathematicians, independently developed the foundations for analytical geometry
• ANALYTIC GEOMETRY • A branch of mathematics which uses algebraic equations to describe the size and position of geometric figures on a coordinate system.
• ANALYTIC GEOMETRY • The link between algebra and geometry was made possible by the development of a coordinate system which allowed geometric ideas, such as point and line, to be described in algebraic terms like real numbers and equations. • Also known as Cartesian geometry or coordinate geometry.
• ANALYTIC GEOMETRY • The use of a coordinate system to relate geometric points to real numbers is the central idea of analytic geometry. • By defining each point with a unique set of real numbers, geometric figures such as lines, circles, and conics can be described with algebraic equations.
• CARTESIAN PLANE • The Cartesian plane, the basis of analytic geometry, allows algebraic equations to be graphically represented, in a process called graphing. • It is actually the graphical representation of an algebraic equation, of any form -- graphs of polynomials, rational functions, conic sections, hyperbolas, exponential and logarithmic functions, trigonometric functions, and even vectors.
• CARTESIAN PLANE • x-axis (horizontal axis) where the x values are plotted along. • y-axis (vertical axis) where the y values are plotted along. • origin, symbolized by 0, marks the value of 0 of both axes • coordinates are given in the form (x,y) and is used to represent different points on the plane.
• INCLINATION OF A LINE • The smallest angle θ, greater than or equal to 0°, that the line makes with the positive direction of the xaxis (0° ≤ θ < 180°) • Inclination of a horizontal line is 0. • The tangent of the inclination m = tan θ
• INCLINATION OF A LINE y y L L θ O M θ x O M x
• ANGLE BETWEEN TWO LINES
• ANGLE BETWEEN TWO LINES • If θ is angle, measured counter clockwise, between two lines, then • where m2 is the slope of the terminal side and m1 is the slope of the initial side
• A point is an ordered pair of numbers written as (x; y). ● Distance is a measure of the length between two points. ● The formula for finding the distance between any two points is: ●
• The formula for finding the mid-point between two points is:
• SLOPE OF A LINE • The slope or gradient of a line describes the steepness, incline or grade. • A higher slope value indicates a steeper incline. • Slope is normally described by the ratio of the “rise” divided by the “run” between two points on a line. • The slope is denoted by 𝒎.
• Gradient between two points The gradient between two points is determined by the ratio of vertical ● change to horizontal change. The formula for finding the gradient of a line is: ●
• SLOPE OF A LINE • If line rises from left to right, • If line goes from right to left, • If line is parallel to x-axis, •If line is parallel to y-axis, 𝒎> 𝒐 𝒎< 𝟎 𝒎= 𝟎 𝒎 = 𝒖𝒏𝒅𝒆𝒇𝒊𝒏𝒆𝒅
• • Two non-vertical lines are parallel if, and only if, their slopes are equal. • Two slant lines are perpendicular if, and only if, the slope of one is the negative reciprocal of the slope of the other..(If two lines are perpendicular, the product of their gradients is equal to −1.) • For horizontal lines the gradient is equal to 0. • For vertical lines the gradient is undefined.
• If 2 lines with gradients m1 and m2 are perpendicular then m1 × m2 = -1 Conversely: If m1 × m2 = -1 then the two lines with gradients m1 and m2 are perpendicular.
• STRAIGHT LINE FACTS y = mx + c y2 - y1 Gradient = x2 - x1 Y – axis Intercept
• Find the equation of the line which passes through the point (-1, 3) and is perpendicular to the line with equation 4x  y 1  0
• SOLUTION Find gradient of given line: 4 x  y  1  0  y  4 x  1  m  4 Find gradient of perpendicular: Find equation: m 1 (using formula m  m  1) 1 2 4 y – b = m(x – a) y – 3 = ¼ (x –(-1)) 4y – 12 = x + 1 4 y  x  13  0
• This presentation is a mash up of 6 different sources. These are: Felipe, N, M. (2014). Analytical geometry basic concepts [PowerPoint Presentation]. Available at: http://www.slideshare.net/NancyFelipe1/analyticgeometry-basic-concepts. Accessed on: 6 March 2014. Demirdag, D. (2013). Lecture #4 analytic geometry [PowerPoint Presentation]. Available at: http://www.slideshare.net/denmarmarasigan/lecture-4-analytic-geometry. Accessed on: 6 March 2014. Share, S. (2014). Analytical geometry [PowerPoint Presentation]. Available at: http://www.slideshare.net/SuziShare/analytical-geometry. Accessed on: 6 March 2014. Derirdag, D. (2012).Analytical geometry [PowerPoint Presentation]. Available at: http://www.slideshare.net/mstfdemirdag/analytic-geometry. Accessed on: 6 March 2014. Nolasco, C, M. Analytical geometry [PowerPoint Presentation]. Available at: http://www.slideshare.net/CecilleMaeNolasco/analytical-geometry. Accessed on: 6 March 2014. Siyavula_Education. (2012).Analytical geometry Everything Maths, Grade 10 [PowerPoint Presentation}. Available at: http://www.slideshare.net/Siyavula_Education/analyticalgeometry?qid=9526d5d1-098f-45da-90b6d503de7f2db5&v=default&b=&from_search=4). Accessed on: 6 March 2014.