Question

Find the locus of a point which moves so that the difference of its distances from two fixed straight lines at right angles is equal to its distance from a fixed straight line.

Answer 1

Let the point be $P(h,k)$ and two perpendicular fixed straight lines be

$ax+by+c=\mathrm{0.......}\left(i\right)bx-ay+d=\mathrm{0........}\left(ii\right)$

and the third fixed straight line be

$Ax+By+C=\mathrm{0........}\left(iii\right)$

Perpendicular distance from $P$ to $\left(i\right)$ is

$p_1=\frac{ah+bk+c}{\sqrt{a^2+b^2}}$

From $\left(ii\right)$

$p_2=\frac{bh-ak+d}{\sqrt{b^2+a^2}}$

From $\left(iii\right)$

$p_3=\frac{Ah+Bk+C}{\sqrt{A^2+B^2}}$

Given $p_1-p_2=p_3$

$\frac{ah+bk+c}{\sqrt{a^2+b^2}}-\frac{bh-ak+d}{\sqrt{b^2+a^2}}=\frac{Ah+Bk+C}{\sqrt{A^2+B^2}}\frac{ah+bk+c-bh+ak-d}{\sqrt{a^2+b^2}}=\frac{Ah+Bk+C}{\sqrt{A^2+B^2}}\frac{(a-b)h+(a+b)k+(c-d)}{\sqrt{a^2+b^2}}=\frac{Ah+Bk+C}{\sqrt{A^2+B^2}}\sqrt{A^2+B^2}(a-b)h+\sqrt{A^2+B^2}(a+b)k+\sqrt{A^2+B^2}(c-d)=A\sqrt{a^2+b^2}h+B\sqrt{a^2+b^2}k+C\sqrt{a^2+b^2}$

Replacing $h$ by $x$ and $k$ by $y$

$\left(\sqrt{A^2+B^2}\right(a-b)-A\sqrt{a^2+b^2})x+\left(\sqrt{A^2+B^2}\right(a+b)-B\sqrt{a^2+b^2})y+\sqrt{A^2+B^2}(c-d)-C\sqrt{a^2+b^2}=0Dx+Ey+F=0$

which is also a straight line . So the locus of $P$ is a straight line