**Example 4**

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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 = β((π₯2 βπ₯1)2+(π¦2 βπ¦1)2) = β((7 βπ₯)2+(1 βπ¦)2) Similarly, finding AC x1 = x , y1 = y x2 = 3 , y2 = 5 AC = β((π₯2 βπ₯1)2+(π¦2 βπ¦1)2) = β(( 3 βπ₯)2+(5 βπ¦)2) We know that AB = AC β(( 7 βπ₯)2+(1 βπ¦)2) = β(( 3 βπ₯)2+(5 βπ¦)2) Squaring both sides (β(( 7 βπ₯)2+(1 βπ¦)2))2 = (β(( 3 βπ₯)2+(5 βπ¦)2) ) 2 (7 β x)2 + (1 β y)2 = (3 β x) 2 + (5 β y) 2 72 + x2 β 2 (7) (x) + 12 + y2 β 2y = (3) 2 + x2 β 2 (3)x + 52 + y2 β 2 (5)y 49 + x2 β 14x + 1 + y2 β2y = 9 + x2 β 6x + 25 + y2 β 10y (x2 β x2) β 14x + 6x + y2 β y2 β2y + 10y + 9 β 49 β 1 + 25 = 0 β8x + 8y + 16 = 0 8(βx + y + 2) = 0 βx + y + 2 = 0/8 βx + 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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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.