Rational Solver
0 votes
44 views
If the sum of the distances of a point from two perpendicular lines in a plane is 1 unit, then its locus is

(A) a square (B) a circle (C) a straight line (D) two intersecting lines
in WBJEE2022 by Professor
edited by | 44 views

1 Answer

0 votes
Best answer

Let P(h,k) be the point that satisfies the given property. We first consider the perpendicular lines. Two most simple perpendicular lines are 

$y=0$ (that is, $x$-axis) and $x=0$ ($y$-axis).


Recall that the distance between a point $(h,k)$ and a line $ax+by+c=0$ is given by 

$$ D = \left|\dfrac{ah+bk+c}{\sqrt{a^2+b^2}}\right|$$


Now the distance between the point $P$ and $x$-axis is $|k|$.

The distance between the point $P$ and $y$-axis is $|h|$. 

It is given the sum of the above two distances is one. That means, 

$$|h|+|k|=1$$


Note that the above equation is a set of four equations, which are

  • $h+k=1$, when $P(h,k)$ lies on the first quadrant.
  • $-h+k=1$, when $P(h,k)$ lies on the second quadrant.
  • $-h-k=1$, when $P(h,k)$ lies on the third quadrant.
  • $h-k=1$, when $P(h,k)$ lies on the fourth quadrant.

Note that above four equation represents together a square with corners at $(\pm1,0)$, $(0,\pm 1)$.


The correct answer is $A$. 

by Professor
edited by
Thank You for providing an answer to this question.
Welcome to Rational Solver, where you can ask questions and receive answers from other members of the community.
65 questions
23 answers
2 comments
1,801 users