Q.Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).
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1 thought on “Q.Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).<br />.<br />.<br />”

1. Answer:

   Given : point (x, y) is equidistant from the point(7,1) and (3,5)​

   To Find : relation between x and y

   Solution:

   point (x, y) is equidistant from the point(7,1) and (3,5)​

   (x – 7)² + ( y – 1)² = (x – 3)² + ( y – 5)²

   => x² – 14x + 49 + y² – 2y + 1 = x² – 6x + 9 + y² -10y + 25

   => -14x – 2y + 50 = -6x – 10y + 34

   => 8x – 8y = 16

   => x – y = 2

   Another way :

   point (x, y) is equidistant from the point (7,1) and (3,5)​

   hence its perpendicular bisector

   mid point = ( 7 +3)/2 , ( 1 + 5)/2

   = 5 , 3

   Slope between points = ( 3 – 1)/( 5 – 7) = – 1

   Hence Slope of perpendicular line = 1

   y – 3 = 1 (x – 5)

   => y – 3 = x – 5

   => x – y = 2

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