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" I think, therefore I am .". René Descartes. 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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"I think, therefore I am." René Descartes 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.
René Des ‘cartes’ “Cartes”ian 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 toLa Géometrieand Descartes's invention of coordinate geometry.
Algebraic Equation in Geometry x – 2y = 1 Line Geogebra X—2y = 1 is a line in Geometry
Turning point in the History of Mathematics After 2000 years of Euclidean Geometry This was the FIRSTsignificant development by RENE DESCARTES ( French) in 17th Century, Part of the credit goes to Pierre Fermat’s (French) pioneering work in analytic geometry. Sir Isaac Newton(1640–1727) developedtendifferent coordinate systems. It was Swiss mathematician Jakob Bernoulli (1654–1705) who first used a polar co-ordinate system for calculus Newton and Leibnitz used the polar coordinate system
┴ • 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
Use of Co-ordinate Geometry Cell Address is (D,3) or D3
Use of Co-ordinate Geometry R A D A R MAP R A D A R
Use of Co-ordinate Geometry Each Pixel uses x-y co-ordinates Pixels in Digital Photos
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..
The screenyou are looking at is a grid of thousands of tiny dots called pixels that together make up the image
Practical Application: All computer programs written in Java language, uses distance between two points.
Frame of reference Vertical II Half Plane I Above X-Axis Horizontal Left of Y-axis origin Right of Y-axis Terms Abscissa Ordinate Ordered Pair Quadrants Sign –Convention IV Below X-Axis III
z y a 0 b y x x Dimensions • 1-D • 2-D • 3-D
a 0 b Distance Formula 1-D • | b-a | or • | a-b |
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!!
Distance between two points.In general, y B(x2,y2) AB2 = (y2-y1)2 + (x2-x1)2 y2 Hence, the formula for Length of AB or Distance between A and B is Length = y2 – y1 y1 A(x1,y1) Length = x2 – x1 x x1 x2
Distance between two points. X2 - x1 = 18-5 A ( 5 , 3 ) , B ( 18, 17 ) A ( x1 , y1 ) B ( x2 , y2 ) y2 - y1 = 17-3 y Using Pythagoras’ Theorem, AB2 = (18 - 5)2 + (17 - 3)2 B(18,17) 17 AB2 = 132 + 142 17 – 3 = 14 units 3 A(5,3) 18 – 5 = 13 units x 5 18
Distance formula is nothing but Pythagoras Theorem B A
The mid-point of two points. Look at it’s horizontal length y B(18,17) Mid-point of AB y2 Look at it’s vertical length Formula for mid-point is y1 A(5,3) x x1 x2
10 11.5 The mid-point of two points. Look at it’s horizontal length y Mid-point of AB B(18,17) 17 = 11.5 (11.5, 10) Look at it’s vertical length 3 A(5,3) (18,3) = 10 x 5 18
Find the distance between the points (-1,3) and (2,-6) y2—y1= -6-3= -9 x2—x1=2--(--1)= 3 (-1, 3) (2, -6) (x1 , y1 ) (x2 ,y2 ) AB= 9.49 units (3 sig. fig)