Q6 of 49 Page 1

Find a relation between x and y such that the point P(x, y) is equidistant from the points A (2, 5) and B (-3, 7).

Given that AP=BP

Distance between two points (x1,y1) and (x2,y2) is given by


√ (x1-x2)2 + (y1-y2)2


Length of AP=√ (x-2)2 + (y-5)2


Length of BP=√ (x + 3)2 + (y-7)2


AP=BP


√ (x-2)2 + (y-5)2 =√ (x + 3)2 + (y-7)2


Squaring both sides,


⇒ (x-2)2 + (y-5)2 = (x + 3)2 + (y-7)2


⇒ (x2 + 4-4x) + (y2 + 25-10y) = (x2 + 9 + 6x) + (y2 + 49-14y)


⇒ 29-4x-10y=58 + 6x-14y


⇒ 10x + 29=4y


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