Question

P
S. If point (x, y) is equidistant from the points (a + b, b - a) and
(a - b, a+b) then prove bx = ay.
(HOTS)​

Answer 1

Answer:

Step-by-step explanation:

Let P(x,y) A(a+b,b-a) B(a-b,a+b)

According to question PA=PB [Since they are equidistant]

Using distance formula

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

=√[(a+b)-x]∧2+[(b-a)-y]∧2

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

=$\sqrt{ a^{2}+ b^{2}+2ab+x^{2} -2ax-2bx+b^{2}+a^{2}-2ba+y^{2}-2by+2ay$

=$\sqrt{2a^{2}+2b^{2}+x^{2}-2ax-2bx+y^{2}-2by+2ay }$-----------Eq.1

PB=$\sqrt{ [ (a-b)-x)^{2}+[(a+b)-y]^{2}$

=$\sqrt{ [(a-b)^{2}+x^{2}-2(a-b)(x)]+[(a+b)^{2}+y^{2}-2(a+b)(y)]$

=$\sqrt{ a^{2}+b^{2}-2ab+x^{2}-2ax+2bx+a^{2}+b^{2}+2ab+y^{2}-2ay-2by$

=$\sqrt{ 2a^{2}+2b^{2}+x^{2}-2ax+2bx+y^{2}-2ay-2by$---------Eq.2

Equate-1 and 2

$\sqrt{2a^2+2b^2+x^2-2ax-2bx+y^2-2by+2ay$=$\sqrt{2a^2+2b^2+x^2-2ax+2bx+y^2-2ay-2by$

Simplifying it

$2a^2+2b^2+x^2-2ax-2bx+y^2-2by+2ay$=$2a^2+2b^2+x^2-2ax+2bx+y^2-2ay-2by$

$-2bx+2ay=2bx-2ay\\-2bx-2bx=-2ay-2ay\\-4bx=-4ay\\4ay=4bx\\ay=bx\\bx=ay$

Hence proved........