René Descartes " I think, therefore I am." Founder of Analytic Geometry Descartes lived during the early 17th century. Descartes found a way to describe

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  • Ren Descartes " I think, therefore I am." Founder of Analytic Geometry Descartes lived during the early 17th century. Descartes found a way to describe curves in an arithmetic way. He developed a new method called coordinate geometry, which was basic for the future development of science.
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  • Ren Des cartes Cartesian Co Ordinate System Geometry and the Fly One morning Descartes noticed a fly walking across the ceiling of his bedroom. As he watched the fly, Descartes began to think of how the fly's path could be described without actually tracing its path. His further reflections about describing a path by means of mathematics led to La G ometrie and Descartes's invention of coordinate geometry.
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  • Line X2y = 1 is a line in Geometry
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  • After 2000 years of Euclidean Geometry This was the FIRST significant development by RENE DESCARTES ( French) in 17 th Century, Part of the credit goes to Pierre Fermats (French) pioneering work in analytic geometry. Sir Isaac Newton (16401727) developed ten different coordinate systems. It was Swiss mathematician Jakob Bernoulli (16541705) who first used a polar co-ordinate system for calculus Newton and Leibnitz used the polar coordinate system
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  • Two intersecting line determine a plane. Two intersecting Number lines determine a Co-ordinate Plane/system. or Cartesian Plane. or Rectangular Co-ordinate system. or Two Dimensional orthogonal Co-ordinate System or XY-Plane GRID
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  • Use of Co-ordinate Geometry Cell Address is (D,3) or D3
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  • Use of Co-ordinate Geometry
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  • R A D A R MAP RADARRADAR
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  • Use of Co-ordinate Geometry Each Pixel uses x-y co-ordinates
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  • Coordinate geometry is also applied in scanners. Scanners make use of coordinate geometry to reproduce the exact image of the selected picture in the computer. It manipulates the points of each piece of information in the original documents and reproduces them in soft copy. Thus coordinate geometry is widely used without our knowing..
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  • The screen you are looking at is a grid of thousands of tiny dots called pixels that together make up the image
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  • Practical Application: All computer programs written in Java language, uses distance between two points.
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  • Terms Horizontal Vertical Above X-Axis Below X-Axis Right of Y-axis Left of Y-axis Half Plane origin Abscissa Ordinate Ordered Pair Quadrants Sign Convention I IV II III
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  • Co-ordinate Geometry Statistics Linear Equations Mensuration Graph Construction Trigonometry Points & lines Congruency Similarity Pythagoras Theorem
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  • Dimensions 1-D 2-D 3-D a 0 b y x y x z
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  • 1-D | b-a | or | a-b | a0b
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  • 2-D: THE Distance formula A B
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  • A B
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  • The modern applications of MapQuest, Google Maps, and most recently, GPS devices on phones, use coordinate geometry. Satellites have taken a 3-d world and made it a 2-d grid in which locations have numbers and labels. The GPS system takes these numbers and labels and maps out directions, times and mileage using the satellite given locations to tell you how to get from one place to another, how long it will be and how much time it will take! Amazing!!
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  • Distance between two points. In general, x1x1 x2x2 y1y1 y2y2 A(x 1,y 1 ) B(x 2,y 2 ) Length = x 2 x 1 Length = y 2 y 1 AB 2 = (y 2 -y 1 ) 2 + (x 2 -x 1 ) 2 Hence, the formula for Length of AB or Distance between A and B is y x
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  • Distance between two points. 518 3 17 A(5,3) B(18,17) 18 5 = 13 units 17 3 = 14 units AB 2 = 13 2 + 14 2 Using Pythagoras Theorem, AB 2 = (18 - 5) 2 + (17 - 3) 2 y x A ( 5, 3 ), B ( 18, 17 ) A ( x 1, y 1 ) B ( x 2, y 2 ) y 2 - y 1 = 17-3 X 2 - x 1 = 18-5
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  • A B
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  • The mid-point of two points. x1x1 x2x2 y1y1 A(5,3) B(18,17) Look at its horizontal length Look at its vertical length Mid-point of AB y x y2y2 Formula for mid-point is
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  • The mid-point of two points. 518 3 17 A(5,3) B(18,17) Look at its horizontal length = 11.5 11.5 Look at its vertical length = 10 10 (11.5,10) Mid-point of AB y x (18,3)
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  • Find the distance between the points (-1,3) and (2,-6) (-1, 3) (2, -6) (x 1, y 1 ) (x 2,y 2 ) AB= 9.49 units (3 sig. fig) 2 y 2 y 1 = -6-3= -9 2 x 2 x 1 =2--(--1)= 3