# Example 4 - Chapter 7 Class 10 Coordinate Geometry (Term 1)

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

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Example 4 You are here

Example 5 Important

Example 6

Example 7

Example 8

Example 9 Important

Example 10

Example 11 Deleted for CBSE Board 2022 Exams

Example 12 Deleted for CBSE Board 2022 Exams

Example 13 Deleted for CBSE Board 2022 Exams

Example 14 Important Deleted for CBSE Board 2022 Exams

Example 15 Deleted for CBSE Board 2022 Exams

Chapter 7 Class 10 Coordinate Geometry (Term 1)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.