Question
If the point $$P\left( {x,\,y} \right)$$ is equidistant from the points $$A\left( {a + b,\,b - a} \right)$$ and $$B\left( {a - b,\,a + b} \right)$$ then :
A.
$$ax = by$$
B.
$$bx = ay{\text{ and }}P{\text{ can be }}\left( {a,\,b} \right)$$
C.
$${x^2} - {y^2} = 2\left( {ax + by} \right)$$
D.
None of the above
Answer :
$$bx = ay{\text{ and }}P{\text{ can be }}\left( {a,\,b} \right)$$
Solution :
$$\eqalign{
& {\text{We have, }}PA = PB \cr
& \Rightarrow {\left( {PA} \right)^2} = {\left( {PB} \right)^2} \cr
& \Rightarrow {\left[ {x - \left( {a + b} \right)} \right]^2} + {\left[ {y - \left( {b - a} \right)} \right]^2} = {\left[ {x - \left( {a - b} \right)} \right]^2} + {\left[ {y - \left( {a + b} \right)} \right]^2} \cr
& \Rightarrow {\left[ {\left( {x - a} \right) - b} \right]^2} + {\left[ {\left( {y - b} \right) + a} \right]^2} = {\left[ {\left( {x - a} \right) + b} \right]^2} + {\left[ {\left( {y - b} \right) - a} \right]^2} \cr
& \Rightarrow {\left[ {\left( {x - a} \right) + b} \right]^2} - {\left[ {\left( {x - a} \right) - b} \right]^2} = {\left[ {\left( {y - b} \right) + a} \right]^2} - {\left[ {\left( {y - b} \right) - a} \right]^2} \cr
& \Rightarrow 4b\left( {x - a} \right) = 4a\left( {y - b} \right) \cr
& \Rightarrow bx = ay......\left( {\text{i}} \right) \cr} $$
Also, $$P\left( {a,\,b} \right)$$ satisfies the condition $$\left( {\text{i}} \right)$$ so that $$P$$ can be $$\left( {a,\,b} \right).$$