Question

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

• A. square
• B. circle
• C. straight line
• D. two intersecting line

Answer 1

The correct option is C straight line

Let the point be $(h,k)$

and the two perpendicular lines lines are $ax+by+c=0$ and $bx-ay+d=0$

then, $\frac{ah+bk+c}{\sqrt{a^2+b^2}}+\frac{bh-ak+d}{\sqrt{b^2+a^2}}=1$

$ah+bk+c+bh-ak+d=\sqrt{a^2+b^2}$

$(a+b)h+(b-a)k+c+d-\sqrt{a^2+b^2}=0$

The locus of the point $(h,k)$ is $(a+b)x+(b-a)y+c+d-\sqrt{a^2+b^2}=0$

which is straight line.