Q11 of 184 Page 10

If the point (x, y) on the tangent is equidistant from the points (2, 3) and (6, - 1), find the relation between x and y.

Given points are A(2, 3) and B(6, - 1). It is told that S(x, y) is equidistant from A and B.



So, we get SA = SB,


We know that distance between two points (x1, y1) and (x2, y2) is .


Now,


⇒ SA = SB


⇒ SA2 = SB2


⇒ (x - 2)2 + (y - 3)2 = (x - 6)2 + (y - (- 1))2


⇒ (x - 2)2 + (y - 3)2 = (x - 6)2 + (y + 1)2


⇒ x2 - 4x + 4 + y2 - 6y + 9 = x2 - 12x + 36 + y2 + 2y + 1


⇒ 8x - 8y = 24


⇒ x - y = 3


∴ The relation between x and y is x - y = 3.


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