Question

A straight line $L$ through the origin meets the line $x+y=1$ and $x+y=3$ at $P$ and $Q$ respectively.Through $P$ and $Q$ two straight lines $L_1$ and $L_2$ aredrawn, parallel to $2x-y=5$ and $3x+y=5$respectively. Lines $L_1$ and $L_2$ intersect at $R$. Show that the locus of $R$, as $L$ varies, is a straight line.

Answer 1

Any line through origin is $y=mx$ (m variable). It meets the given lines in points.

$P(\frac{1}{m+1},\frac{m}{m+1})$ and $Q(\frac{3}{m+1},\frac{3m}{m+1})$

Any line $L_1$ through $P$ parallel to $2x-y=5$ is

$2(x-\frac{1}{m+1})-(y-\frac{m}{m+1})=0$

Or $2x-y=\frac{2-m}{m+1}$

Any line through $Q$ parallel to $3x+y=5$ is

$3(x-\frac{3}{m+1})+(y-\frac{3m}{m+1})=0$

Or $3x+y=\frac{3(3+m)}{m+1}$

In order to find the locus of their point of intersection, we have to eliminate the variable $m$ between their equations. We rewrite the equations of lines as

$2x-y=\frac{-m-1+3}{m+1}$ or $2x-y+1=\frac{3}{m+1}$

$3x+y=\frac{3(2+m+1)}{m+1}$ or $3x-y-3=\frac{6}{m+1}$

Dividing the above two thereby eliminating the variable $m$, the required locus is given by

$\frac{2x-y+1}{3x+y-3}=\frac{1}{2}$

Or $4x-2y+2=3x+y-3$

Or $x-3y+5=0$

Which being of first degree, represents the equation of a line.