Question

A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines $ L_1 : 2x + y + 6 = 0 $ and $ L_2 : 4x + 2y - p = 0 $, $ p>0 $, at the points A and B, respectively. If $ AB = \frac{9}{\sqrt{2}} $ and the foot of the perpendicular from the point A on the line $ L_2 $ is M, then $ \frac{AM}{BM} $ is equal to

• A. 5
• B. 4
• C. 2
• D. 3

Hint:

Use the distance formula and the formula for the angle between two lines to solve the problem.

Answer 1

The Correct Option is D

The line passing through the origin and making equal angles with the positive coordinate axes is \( y = x \).
To find the coordinates of A, we solve \( 2x + y + 6 = 0 \) and \( y = x \): \( 2x + x + 6 = 0 \) \( 3x = -6 \) \( x = -2 \) \( y = -2 \) So, A is \( (-2, -2) \). 
To find the coordinates of B, we solve \( 4x + 2y - p = 0 \) and \( y = x \): \( 4x + 2x - p = 0 \) \( 6x = p \) \( x = \frac{p}{6} \) \( y = \frac{p}{6} \) 
So, B is \( \left( \frac{p}{6}, \frac{p}{6} \right) \). 

Given \( AB = \frac{9}{\sqrt{2}} \), we have: \( \sqrt{ \left( \frac{p}{6} + 2 \right)^2 + \left( \frac{p}{6} + 2 \right)^2 } = \frac{9}{\sqrt{2}} \) \( \sqrt{ 2 \left( \frac{p}{6} + 2 \right)^2 } = \frac{9}{\sqrt{2}} \) \( \sqrt{2} \left| \frac{p}{6} + 2 \right| = \frac{9}{\sqrt{2}} \) \( \left| \frac{p}{6} + 2 \right| = \frac{9}{2} \)

Since \( p > 0 \), \( \frac{p}{6} + 2 = \frac{9}{2} \) \( \frac{p}{6} = \frac{9}{2} - 2 = \frac{5}{2} \) \( p = 15 \) 

Therefore, B is \( \left( \frac{15}{6}, \frac{15}{6} \right) = \left( \frac{5}{2}, \frac{5}{2} \right) \). 

The slope of \( L_2 \) is \( m_2 = -2 \). The slope of \( y = x \) is \( m_1 = 1 \). 

Let \( \theta \) be the angle between the lines \( y = x \) and \( L_2 \). \( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{1 - (-2)}{1 + 1(-2)} \right| = \left| \frac{3}{-1} \right| = 3 \) From the geometry, \( \tan \theta = \frac{AM}{BM} \). 

Therefore, \( \frac{AM}{BM} = 3 \).