Question

Let $y=mx+{\lambda }_i,i=1,2,3,...,n$ be a family of $n$ parallel lines subjected to following conditions.
$\left(1\right)m$ being a constant
$\left(2\right)\sum _{i=1}^n{\lambda }_i=1$
A variable line through origin intersects the lines at $P_i(i=1,2,3,...,n)$ and $Q$ be a point on variable line such that $\sum _{i=1}^{n}O{P}_i=OQ$. If the locus of $Q$ is a straight line which passes through a fixed point $(a,b)\forall m\in R,$ then the value of $(3a+2b)$ is

• A. $1$
• B. $2$
• C. $5$
• D. $10$

Answer 1

The correct option is B $2$


Let the variable line through origin $(0,0)$ is $y=x\tan \theta$ and point $Q$ on it is $(x,y)$ where $x=r\cos \theta$ and $y=r\sin \theta$
On solving the lines $y=x\tan \theta$ with $y=mx+{\lambda }_1,$ we get
$(\tan \theta -m)x={\lambda }_1$
$\therefore x=\frac{{\lambda }_1}{\tan \theta -m}=r_1\cos \theta$
$\Rightarrow r_1=\frac{{\lambda }_1}{\sin \theta -m\cos \theta }$
$\Rightarrow r_i=\frac{{\lambda }_i}{\sin \theta -m\cos \theta }$

Also, $r_1+r_2+...+r_n=OQ=r$
$\Rightarrow \frac{{\lambda }_1+{\lambda }_2+...+{\lambda }_n}{\sin \theta -m\cos \theta }=r$
Since, $\sum _{i=1}^n{\lambda }_i=1$
$\therefore 1=r\sin \theta -mr\cos \theta$
$\Rightarrow 1=y-mx$
$\Rightarrow y=mx+1$
which is the locus of a straight line
As, above line passes through $(0,1)\forall m\in R,$ so $(3a+2b)=2$