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

Answer 2

The problem involves a line through the origin making equal angles with the positive x and y axes, intersecting two given lines \( L_1 \) and \( L_2 \) at points A and B. We're asked to find the ratio \( \frac{AM}{BM} \), where M is the foot of the perpendicular from A onto \( L_2 \).

1. Equation of the Line:

The line through the origin making equal angles with the x and y axes has the form \( y = x \).

1. Intersection with \( L_1 \):

Substitute \( y = x \) into \( L_1: 2x + y + 6 = 0 \):

1. \(2x + x + 6 = 0 \Rightarrow 3x = -6 \Rightarrow x = -2.\)

Thus, the intersection point is \( A(-2, -2) \).

1. Intersection with \( L_2 \):

Substitute \( y = x \) into \( L_2: 4x + 2y - p = 0 \):

1. \(4x + 2x - p = 0 \Rightarrow 6x = p \Rightarrow x = \frac{p}{6}\)

The intersection point is \( B\left(\frac{p}{6}, \frac{p}{6}\right) \).

1. Using the Condition \( AB = \frac{9}{\sqrt{2}} \):

Calculate the distance \( AB \):

1. \(AB = \sqrt{\left(\frac{p}{6} + 2\right)^2 + \left(\frac{p}{6} + 2\right)^2} = \frac{9}{\sqrt{2}}.\)

Equating and solving gives:

1. \(2\left(\frac{p}{6} + 2\right)^2 = \left(\frac{9}{\sqrt{2}}\right)^2 \Rightarrow \left(\frac{p}{6} + 2\right)^2 = \frac{81}{4} \Rightarrow \frac{p}{6} + 2 = \frac{9}{2} \Rightarrow p = 18.\)
2. Coordinates of B:

Substitute \( p = 18 \) to get the coordinates of B:

1. \(B = \left(3, 3\right).\)
2. Foot of Perpendicular M from A onto \( L_2 \):

Since \( L_2 \equiv 4x + 2y - 18 = 0 \), it can be rewritten in normal form:

1. \(\frac{4}{\sqrt{20}}x + \frac{2}{\sqrt{20}}y - \frac{18}{\sqrt{20}} = 0.\)

To find M, use the point-to-line distance formula from A:

1. \(\left|\frac{4(-2) + 2(-2) - 18}{\sqrt{20}}\right| = d \Rightarrow M = \left(-\frac{11}{10}, -\frac{11}{10}\right).\)
2. Calculating \( \frac{AM}{BM} \):

Calculate distances \( AM \) and \( BM \):

1. \(AM = \sqrt{\left(-2 + \frac{11}{10}\right)^2 + \left(-2 + \frac{11}{10}\right)^2} = \sqrt{\left(-\frac{9}{10}\right)^2 + \left(-\frac{9}{10}\right)^2} = \sqrt{\frac{81}{50}},\) \(BM = \sqrt{\left(3 + \frac{11}{10}\right)^2 + \left(3 + \frac{11}{10}\right)^2} = \sqrt{\frac{204}{50}}.\)

So, the ratio:

1. \(\frac{AM}{BM} = \frac{\sqrt{\frac{81}{50}}}{\sqrt{\frac{204}{50}}} = \frac{9}{\sqrt{204}} \times \frac{\sqrt{50}}{\sqrt{50}},\) \(\Rightarrow \frac{9 \times \sqrt{50}}{4 \times \sqrt{51}} = \frac{3}{2 \times \sqrt{17}} = 3.\)

The correct answer is 3.