First slide

Equation of line in Straight lines

Question

A straight line through the origin $O$ meet the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points P and Q respectively. The point $O$ divides the segment $P Q$ in the ratio

Moderate

A

1 : 2
B

3 : 4
C

2 : 1
D

4 : 3

Solution

It is clear that the lines lie on opposite side of the origin $O$ . Let the equation of any line through $O$ be $\frac{x}{\cos θ} = \frac{y}{\sin θ} . If O P = r_{1}$ If $O P = r_{1} and O Q = r_{2}$ then the coordinates of $p$ are $(r_{1} \cos θ, r_{1} \sin θ)$ and that of $Q$ are
$(- r_{2} \cos θ, - r_{2} \sin θ)$
Since $P$ lies on $4 x + 2 y = 9, 2 r_{1} (2 \cos θ + \sin θ) = 9$
and $Q$ lies on $2 x + y + 6 = 0, - r_{2} (2 \cos θ + \sin θ) = - 6$
so that $\frac{r_{1}}{r_{2}} = \frac{9}{12} = \frac{3}{4}$
and the required ratio is thus 3 : 4
Alternately Let the equation of the line through $O$ be $y = m x$ ,
then coordinates of $P$ and $Q$ are
respectively $(\frac{9}{4 + 2 m}, \frac{9 m}{4 + 2 m}) and (\frac{- 6}{2 + m}, \frac{- 6 m}{2 + m})$ so that
$\frac{O P}{O Q} = \frac{9}{| 4 + 2 m |} \times \frac{| 2 + m |}{6} = \frac{3}{4}$

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