# Deducing the locus of a point of intersection of two lines.

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 $$R$$.

## 1 Answer

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.

• I did not understand what you meant by, "the main idea is NOT to try to find a value of $\lambda$...", why is this so? Also, why does eliminating $m$ give the required equation? The problem states that the locus is a straight line that is why I assumed it to be... – JC2000 Apr 1 at 6:14
• @JC2000 The question is asking you to SHOW that it is a straight line. It is something you have to PROVE. So, even though the question says it is straight, you CANNOT assume it is straight. You also cannot use the formula $L_1+\lambda L_2=0$, because this is for finding the equation of a straight line, and you cannot use $L_1+\lambda L_2=0$ until you have PROVED that the locus is a straight line. – Jethro Apr 1 at 6:18
• I see, now I understand why you cannot use that particular equation. Could you also explain why eliminating $m$ gives the required equation? – JC2000 Apr 1 at 6:21
• @JC2000 I have shown you how to do this in my answer – Jethro Apr 1 at 6:22
• @JC2000 Yes you are right – Jethro Apr 1 at 13:03