Question

find a relation between x and y such that the points x,y is equidistant from the points (3,6) and (-3,4)​

Answer 1

Let the Points be =>

=> A ( x , y )

=> B ( 7 , 1 )

=> C ( 3, 5 )

♦According to the given question ->

=> AB = AC

◾Applying Distance Formula ,

=> √ ( x2 - x1 )² + ( y2 - y1 )²

♦Distance AB =>

=> √ ( 7 - x )² + ( 1 - y )²

=> √ 49 + x² - 14x + 1 + y² - 2y

=> √ x² + y² + 50 - 14x - 2y .....................(1)

♦Distance AC =>

=> √ ( x - 3)² + ( y - 5 )²

=> √ x² + 9 - 6x + y² + 25 - 10y

=> √ x² + y² + 36 - 6x - 10 y.....................(2)

◾Putting (1 ) and (2 ) Equal & Squaring both Sides ,

=> x² + y² + 36 - 6x - 10y = x² + y² + 50 - 14x -2y

=> -8x + 8y + 16 = 0

=> -8 ( x - y -2 ) = 0

=> x - y = 2

=> x = y + 2

✔Hence the Relation is x = y + 2