Question

If the point P$(x,y)$ is equidistant from the point $A(a+b,b-a)$ and $B(a-b,a+b)$, then

• A. $ax=by$
• B. $bx=ay$
• C. $x^2-y^2=2(ax+by)$
• D. $P$ can be $(a,b)$.

Answer 1

Answer: (B), (D)

$bx=ay$
$P$ can be $(a,b)$.

We have $PA=PB$, i.e., $(PA{)}^2=(PB{)}^2$
$\Rightarrow {[x-(a+b\left)\right]}^2+{[y-(b-a\left)\right]}^2={[x-(a-b\left)\right]}^2+{[y-(a+b\left)\right]}^2$
$\Rightarrow {\left[\right(x-a)-b]}^2+{\left[\right(y-b)+a]}^2={\left[\right(x-a)+b]}^2+{\left[\right(y-b)-a]}^2$
$\Rightarrow {\left[\right(x-a)+b]}^2-{\left[\right(x-a)-b]}^2=-{\left[\right(y-b)+a]}^2+{\left[\right(y-b)-a]}^2$
$\Rightarrow 4b(x-a)=4a(y-b)\Rightarrow bx=ay$ (1)
Therefore, $\left(b\right)$ is correct. Also, $P(a,b)$ satisfies the condition (1), so that $P$ can be $(a,b)$ and hence $\left(d\right)$ is also correct.