Question

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

Answer 1

Since the point $P(x,y)$ is equidistant from the points $A(a+b,b-a)\equiv (x_1,y_1)$ and $B(a-b,a+b)\equiv (x_2,y_2).$

Using distance formula,

Distance Formula $=\sqrt{{(x-x_1)}^2+{(y-y_1)}^2}$

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

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

Therefore, $PA=PB$

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

Squaring on both the sides,

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

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

$-2x(a+b)-2y(b-a)=-2x(a-b)-2y(b+a)$

$-2ax-2bx-2by+2ay=-2ax+2bx-2by-2ay$

$-4bx=-4ay$

$bx=ay$

Hence proved.