Question

$A$ straight line through the origin $O$ meets the parallel lines $4x+2y=9$ and $2x+y+6=0$ at points $P$ and $Q$ respectively, then point $O$ divides the segment $PQ$ in the ratio:

• A. $1:2$
• B. $3:4$
• C. $2:1$
• D. $4:3$

Answer 1

Answer: (B)

$3:4$

Let, equation of line passing through origin is $y=mx$ --- (1)

$4x+2y=9$ ---(2)

$2x+y=-6$ ---(3)

From intersection of line 1 with line 2

$x_1=\frac{9}{4+2m}$ & $y_1=\frac{9m}{4+2m}$ $P(\frac{9}{4+2m},\frac{9m}{4+2m})$

From intersection of line 1 with line 3

$x_2=\frac{-6}{2+m}$ & $y_2=\frac{-6}{2+m}$ $Q(\frac{-6}{2-m},\frac{-6m}{2+m})$

Let o divides PQ in ration 1 : K

So, by formula

$o=\frac{k\left(\frac{9}{4+2m}\right)-\frac{6}{2+m}}{k+1}$ & $o=\frac{k\left(\frac{9m}{4+2m}\right)-\frac{6m}{2+1}}{k+1}$

$k=\frac{6\times 2}{9}=\frac{4}{3}$

$\therefore k=\frac{4}{3}$

$\therefore ration\Rightarrow 1:k$

$1:\frac{4}{3}$

$\Rightarrow 3:4$

Let, equation of line passing through origin is $y=mx$ --- (1)

$4x+2y=9$ ---(2)

$2x+y=-6$ ---(3)

From intersection of line 1 with line 2

$x_1=\frac{9}{4+2m}$ & $y_1=\frac{9m}{4+2m}$ $P(\frac{9}{4+2m},\frac{9m}{4+2m})$

From intersection of line 1 with line 3

$x_2=\frac{-6}{2+m}$ & $y_2=\frac{-6}{2+m}$ $Q(\frac{-6}{2-m},\frac{-6m}{2+m})$

Let o divides PQ in ration 1 : K

So, by formula

$o=\frac{k\left(\frac{9}{4+2m}\right)-\frac{6}{2+m}}{k+1}$ & $o=\frac{k\left(\frac{9m}{4+2m}\right)-\frac{6m}{2+1}}{k+1}$

$k=\frac{6\times 2}{9}=\frac{4}{3}$

$\therefore k=\frac{4}{3}$

$\therefore ration\Rightarrow 1:k$

$1:\frac{4}{3}$

$\Rightarrow 3:4$