Question

Find the relation between x and y such that the point (x,y) is equidistant from the points (7,1) and (3,5)

Answer 1

Heya!
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♦Coordinate Geometry ♦
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✔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 ✔
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Answer 2

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