Question

If the point p(x y) is equidistant from a(a+b b-a) and b(a-b a+b) prove that bx=ay

Answer 1

Please refer the attachment above

Answer 2

If the point p(x y) is equidistant from a(a+b b-a) and b(a-b a+b), bx = ay proved.

Step-by-step explanation:

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

To find, prove that bx = ay.

∴ $PA^{2} =PB^{2}$

Using distance formula,

$\sqrt{(x_{2}- x_{1})^{2}+(y_{2}- y_{1})^{2}}$

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

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

⇒$(a+b-x+a-b-x)(a+b-x-a+b+x)=(a+b-y+b-a-y)(a+b-y-b+a+y)$

⇒ $(2a-2x)(2b)=(2b-2y)(2a)$

⇒ $(a-x)(b)=(b-y)(a)$

⇒ $ab-bx=ab-ay$

⇒$bx=ay$, proved.