A straight line $L$ through origin meets $x+y=1$...$(1)$ at $P$ and $x+y=3$...$(2)$ at $Q$. Through $P$ and $Q$ two lines $L_1$ and $L_2$ are drawn which are parallel to $2x-y=5$...$(3)$ and $3x-y=5$...$(4)$ respectively. $L_1$ and $L_2$ intersect at $R$. Show that the locus of $R$ as $L$ varies is a straight line.
The solution :
Let $L$ be $y=mx$. Using $(1)$&$(2)$ we get $P$ = $(\frac{1}{m+1},\frac{m}{m+1})$ and $Q$ = $(\frac{3}{m+1},\frac{3m}{m+1})$.
Now, $L_1$ is $y-2x = \frac{m-2}{m+1}$...$(5)$ and $L_2$ is $y+3x=\frac{3m+9}{m+1}...$(6)
Eliminate $m$ from $(5)$ and $(6)$ which gives us the locus of $R$ as $x-3y+5=0$ which represents a straight line.
My Question : I understand that $L_1 + \lambda(L_2) = 0$ represents the set of straight lines passing through the intersection of the two lines (including the line representing the locus of $R$, but I don't understand how the value for $\lambda$ using which $m$ is eliminated also gives us the equation representing the line which is the locus of $R$.
Now, $L_1$ is $y-2x = \frac{m-2}{m+1}$ and $L_2$ is $y+3x=\frac{3m+9}{m+1}$. At the intersection of $L_1$ and $L_2$, $$ 2x+\frac{m-2}{m+1}=-3x+\frac{3m+9}{m+1}. $$ So, $$ 5x=\frac{3m+9}{m+1}-\frac{m-2}{m+1}=\frac{2m+11}{m+1}.\\ 5xm+5x=2m+11\\ (5x-2)m=11-5x\\ m=\frac{11-5x}{5x-2}. $$ Substituting into (5) gives $$ y-2x=\frac{\frac{11-5x}{5x-2}-2}{\frac{11-5x}{5x-2}+1}=\frac{11-5x-10x+4}{11-5x+5x-2}=\frac{15-15x}{9}=\frac{5-5x}{3}\\ 3y-6x=5-5x\\ x-3y+5=0 $$ as required.
The main idea is NOT to try to find a value of $\lambda$ as posted in your question. We can say that the equation of the locus is $L_1 + \lambda(L_2) = 0$ ONLY IF we already know it is a straight line. However, you cannot assume it is a straight line, because the question is asking you to prove that it is a straight line.