# Example 4 - Chapter 7 Class 10 Coordinate Geometry

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 4 Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5). Let the points be A(x, y) , B(7, 1) , C(3, 5) According to question, point A is equidistant from B & C Hence AB = AC So, we will find AB & AC using distance formula Finding AB x1 = x , y1 = y x2 = 7 , y2 = 1 AB = (( 2 1)2+( 2 1)2) = ((7 )2+(1 )2) Similarly, finding AC x1 = x , y1 = y x2 = 3 , y2 = 5 AC = (( 2 1)2+( 2 1)2) = (( 3 )2+(5 )2) We know that AB = AC (( 7 )2+(1 )2) = (( 3 )2+(5 )2) Squaring both sides ( (( 7 )2+(1 )2))2 = ( (( 3 )2+(5 )2) ) 2 (7 x)2 + (1 y)2 = (3 x) 2 + (5 y) 2 72 + x2 2 (7) (x) + 12 + y2 2y = (3) 2 + x2 2 (3)x + 52 + y2 2 (5)y 49 + x2 14x + 1 + y2 2y = 9 + x2 6x + 25 + y2 10y (x2 x2) 14x + 6x + y2 y2 2y + 10y + 9 49 1 + 25 = 0 8x + 8y + 16 = 0 8( x + y + 2) = 0 x + y + 2 = 0/8 x + y + 2 = 0 y + 2 = x x = y + 2 Note that we only have to give relation between x & y and not solve it So, answer is x = y + 2

Chapter 7 Class 10 Coordinate Geometry

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.