Question

A straight line passing through the origin \( O \) meets the parallel lines \( 4x + 2y = 9 \) and \( 2x + y + 6 = 0 \) at the points \( P \) and \( Q \) respectively. Then the point \( O \) divides the line segment \( PQ \) in the ratio

• A. \( 1 : 2 \)
• B. \( 2 : 1 \)
• C. \( 3 : 4 \)
• D. \( 4 : 3 \)

Hint:

To find the ratio in which a point divides a line segment, use the coordinates of intersection and apply distance formula, ensuring consistent direction and signs are considered.

Answer 1

The Correct Option is C

Let the equation of the line through origin be \( y = mx \). It intersects the lines: 1. \( 4x + 2y = 9 \Rightarrow 4x + 2mx = 9 \Rightarrow x(4 + 2m) = 9 \Rightarrow x = \frac{9}{4 + 2m} \) So the point \( P \) is: \[ P = \left( \frac{9}{4 + 2m},\ \frac{9m}{4 + 2m} \right) \] 2. \( 2x + y + 6 = 0 \Rightarrow 2x + mx = -6 \Rightarrow x(2 + m) = -6 \Rightarrow x = \frac{-6}{2 + m} \) So the point \( Q \) is: \[ Q = \left( \frac{-6}{2 + m},\ \frac{-6m}{2 + m} \right) \] Now we find the ratio \( \frac{OP}{OQ} \) using the distances from origin: \[ OP = \sqrt{ \left( \frac{9}{4 + 2m} \right)^2 + \left( \frac{9m}{4 + 2m} \right)^2 } = \frac{9}{4 + 2m} \sqrt{1 + m^2} \] \[ OQ = \sqrt{ \left( \frac{-6}{2 + m} \right)^2 + \left( \frac{-6m}{2 + m} \right)^2 } = \frac{6}{2 + m} \sqrt{1 + m^2} \] Now, \[ \frac{OP}{OQ} = \frac{ \frac{9}{4 + 2m} }{ \frac{6}{2 + m} } = \frac{9(2 + m)}{6(4 + 2m)} = \frac{3(2 + m)}{2(4 + 2m)} \] Simplify: \[ = \frac{3(2 + m)}{2 . 2(2 + m)} = \frac{3}{4} \Rightarrow \text{Ratio } OP : OQ = 3 : 4 \Rightarrow \text{Ratio } PQ \text{ is divided by } O = 3 : 4 \]