Question

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

Answer 1

Given points $P(x,y)\equiv (x_2,y_2)$ ,$A(a+b,b-a)\equiv (x_1,y_1)$

Since point p(x,y) is equidistant from $A(a+b,b-a),B(a-b,a+b)$

$P(x,y)\equiv (x_2,y_2),B(a-b,a+b)\equiv (x_1,y_1)$

$\therefore$ $PA=PB$

$=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

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

Squaring both sides

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

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

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

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

$\Rightarrow bx=ay$ $\left[henceproved\right]$