MCQ
If the point $(x, y)$ be equidistant from the points $(a + b,\,b - a)$ and $(a - b,\,a + b),$ then
  • A
    $ax + by = 0$
  • B
    $ax - by = 0$
  • C
    $bx + ay = 0$
  • $bx - ay = 0$

Answer

Correct option: D.
$bx - ay = 0$
d
(d) ${\left\{ {x - (a + b)} \right\}^2} + {\left\{ {y - (b - a)} \right\}^2} = {\left\{ {x - (a - b)} \right\}^2} + {\left\{ {y - (a + b)} \right\}^2}$

$ \Rightarrow \,\,{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)$

On simplification, we get $bx - ay = 0$

Trick : The locus will be right bisector of the line joining the given points, therefore the line must pass through the mid-points of the given point i.e. $(a, b)$. Obviously, the line given in option $(d)$ passes through $(a, b)$.

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