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Feb. 12, 2020•0 likes•3,363 views

Feb. 12, 2020•0 likes•3,363 views

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Sec 3 E Maths Notes Coordinate Geometry

- 1. 4048 3EM Coordinate Geometry (1) Math Academy® © All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. © www.MathAcademy.sg 1 Be a lifelong student. The more you learn, the more you earn and more self confidence you will have. - Brian Tracy Notes: Coordinate Geometry [A] Distance Formula This formula is an application of Pythagoras' theorem for right triangles: Note that the distance is taken to be positive. Example 1: Given that is an isosceles triangle with vertices , and and , find the value of . Ans: Solution: ABC ( , 1)A p - (2, 5)B (3, 4)C ACAB = p 3- ACAB = Given two points and , the distance between these points is given by the formula: ( )1 1,x y ( )2 2,x y 2 21 2 21 )()( yyxxPQ -+-= Length same, apply distance formula
- 2. 4048 3EM Coordinate Geometry (1) Math Academy® © All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. © www.MathAcademy.sg 2 Example 2: In the diagram below, is a regular hexagon. , and has coordinates , and respectively. (i) State the coordinates of , in terms of . [1] (ii) Justify, showing all workings clearly, why the coordinate of will not be an integer. [2] Solution: (i) (ii) Let the midpoint of be . By Pythagoras theorem, Since 12 is not a perfect square, the coordinate of will not be an integer. Or can apply distance formula to and , whereby . ABCDEF A E F (0,6) ( , 0)s (0,2) B s x - D ( , 8)B s DF X 2 2 2 2 4s + = 12s = 2 12DF = x - D (2 , 2)D s ( , 0)E s 4DE =
- 3. 4048 3EM Coordinate Geometry (1) Math Academy® © All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. © www.MathAcademy.sg 3 Example 6: Given that the gradient of and is , find the possible coordinates of . Ans: Solution: Example 7: The diagram, which is not drawn to scale, shows the three lines , and . (a) Find the coordinates of , and . [4] (b) The point is the same distance from as it is from Find the value of . [1] Ans: (a) (b) 0.5 (3, )A p- 2 ( , )B p p- 2 1 - B ( 1,1), (1.5,2.25)B B- 2 ( ) 1 3 2 p p p - - = - - - 5=y xy -= 3 343 += xy A B C ( ,0)k A B k 6 1 ( 2,5), (3,5), ( ,2 ) 7 7 A B C- Casio mode 3, 3 y x O A B C horizontal Positive gradient Negative gradient
- 4. 4048 3EM Coordinate Geometry (1) Math Academy® © All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. © www.MathAcademy.sg 4 [C] Manipulate the Equation of a straight line Example 8: Determine the gradient and intercept for each of the straight lines in the table below. y - Equation Gradient intercept No need transform No need transform 12 0 No need transform 0 5 4 cmxy += -y xy 3 1 5 -= xy 12= 5=y 182 += xy 1 4 2 y x= + 1 2 105 += yx The equation of a straight line with gradient and intercept ism -y c y mx c= + cmxy += Gradient, must be subjecty
- 5. 4048 3EM Coordinate Geometry (1) Math Academy® © All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. © www.MathAcademy.sg 5 Example 9: The gradient of the line , is . Find the value of , where is a whole number. Ans: Ws 2 [D] Form equation of a straight line (1) Given gradient and pass through a point Example 10: Find the equation of the straight line whose gradient is 3 and passes through the point . Solution: l 2 2 6 0k x ky- - = 9 k k 18k = (4, 2)- y mx c= + 3y x c= + 2 3(4) c- = + 14c = - 3 14y x = - 1. Gradient Equation of Line 2. A point Sub grad 1st Sub point 2nd
- 6. 4048 3EM Coordinate Geometry (1) Math Academy® © All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. © www.MathAcademy.sg 6 (2) Given 2 points Example 11: Find the equation of the line passing through and . Solution: gradient Observe that the intercept is 11 Example 12: The point lies on the line . (a) Find the value of . [1] (b) Find the equation of the line parallel to , passing through the point . [2] Ans: (a) (b) (0,11)A )2,6(B 11 2 3 0 6 2 - = = - - y mx c= + 3 2 y x c= - + y - 3 11 2 y x = - + ( , 2)a 3 2 1 0x y+ - = a 3 2 1 0x y+ - = ( 1, 4)- 1a = - 3 5 2 2 y x= - + Find grad 1st 1. Gradient Equation of Line 2. A point Sub point into equation of line Grad same, manipulate to make the subjecty
- 7. 4048 3EM Coordinate Geometry (1) Math Academy® © All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. © www.MathAcademy.sg 7 Example 13: The diagram shows a regular hexagon, , where and . Given that the length of is units, find (a) the value of . [2] (b) the equation of , [2] Ans: (a) 18 (b) ABCDEF (0,10)B (4, )C p BC 80 p BC 2 10y x= +