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, the 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

Answer 2

The correct answer is B:$3\ratio4$
Given that;
The equation of the line is:-
$4x+2y=9-(i)$ and $2x+y=-6-(ii)$
Let the equation of the line passes through the origin: $y=mx-(iii)$
Then two intersection of (i) and (iii)
$x_1=\frac{9}{4+2m},y=\frac{9m}{4+2m}|p(\frac{9}{4+2m},\frac{9m}{4+2m})$
similarly, (ii) and (iii);
$x_2=\frac{-6}{2+m},y_2=\frac{-6m}{2+m}|q(\frac{-6}{2+m},\frac{-6m}{2+m})$
Let us consider a point ‘O’ denotes PQ in the ratio of $1\ratio{K}$
$\therefore 0=\frac{K(\frac{9}{4+2m}-\frac{6}{2+m})}{K+m}$ and $\frac{K(\frac{9m}{4+2m}-\frac{6m}{2+1})}{K+1}$
$K=\frac{6\times2}{9}=\frac{4}{3}$
$\therefore K=\frac{4}{3}$ $\therefore Ratio\space is \space3\ratio{4}$