If sum of distances of a point from two perpendicular lines in a plane is 1 , then its locus is

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9.Straight Line

medium

If the sum of the distances of a point from two perpendicular lines in a plane is $1$, then its locus is

Square

Circle

Straight line

Two intersecting lines

(IIT-1992)

Solution

(a) Required locus of the point $(x,y)$ is the curve $|x| + |y| = 1$. If the point lies in the first quadrant, then $x > 0,y > 0$ and so $|x| + |y| = 1 \Rightarrow x + y = 1$, which is straight line $AB$. If the point $(x,\,y)$lies in second quadrant then $x < 0$, $y > 0$ and so $|x| + |y| = 1$ ==> $ – x + y = 1$

Similarly for third and fourth quadrant, the equations are $ – x – y = 1$and $x – y = 1$.
Hence the required locus is the curve consisting of the sides of the square $ABCD.$

Standard 11

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Show that the area of the triangle formed by the lines

$y=m_{1} x+c_{1}, y=m_{2} x+c_{2}$ and $x=0$ is $\frac{\left(c_{1}-c_{2}\right)^{2}}{2\left|m_{1}-m_{2}\right|}$.