Question

If the point P(x, y) is equidistant from the points A$(a+b,b-a)$ and B$(a-b,a+b)$, then which of the following condition is true?

• 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, (B) is correct. Also, P(a, b) satisfies the condition (1), so that P can be (a, b) and hence (D) is
also correct.