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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

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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
Thank You for providing an answer to this question.