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If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b). Prove that bx = ay.

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#### Solution

P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b).

∴ AP = BP

∴ `sqrt([x-(a+b)]^2+[y-(b-a)]^2)=sqrt([x-(a-b)]^2+[y-(a+b)]^2`

∴ [x-(a+b)]^{2}+[y-(b-a)]^{2} = [x-(a-b)]^{2}+[y-(a+b)]^{2}

∴ x^{2}-2x(a+b)+(a+b)^{2}+y^{2}-2y(b-a)+(b-a)^{2}

= x^{2}-2x(a-b)+(a-b)^{2}+y^{2}-2y(a+b)+(a+b)^{2}

∴ -2x(a+b)-2y(b-a)=-2x(a-b)-2y(a+b)

∴ ax+bx+by-ay=ax-bx+ay+by

∴ 2bx=2ay

∴bx=ay ...(proved)

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