Question

If a line L passing through a point A(2,3) intersects another line \( 4x-3y-19=0 \) at the point B such that \( AB=4 \), then the angle made by the line L with positive X-axis in anti-clockwise direction is

• A. \( \tan^{-1}\left(-\frac{3}{4}\right) \)
• B. \( \tan^{-1}\left(\frac{3}{4}\right) \)
• C. \( \frac{\pi}{4} \)
• D. \( -\frac{\pi}{4} \)

Hint:

If the given distance equals the calculated perpendicular distance, the line segment is normal to the target line. If the given distance is greater, there would be two possible lines (secants).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given the length of the segment AB where A is a fixed point and B lies on a given line. Checking the perpendicular distance from A to the line can reveal if AB is the perpendicular dropped from A, which fixes the orientation of line L.
Step 2: Key Formula or Approach:
Perpendicular distance \( d = \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}} \).
Step 3: Detailed Explanation:
Point \( A(2,3) \), Line \( 4x - 3y - 19 = 0 \). Calculate the perpendicular distance \( p \) from A to the line: \[ p = \frac{|4(2) - 3(3) - 19|}{\sqrt{4^2 + (-3)^2}} = \frac{|8 - 9 - 19|}{\sqrt{16+9}} = \frac{|-20|}{5} = 4 \] We are given that the length \( AB = 4 \). Since the distance from point A to the line is exactly equal to the length of the segment AB, the segment AB must be perpendicular to the line \( 4x - 3y - 19 = 0 \). Thus, line L (containing AB) is perpendicular to the given line. Slope of given line \( m_1 = -\frac{4}{-3} = \frac{4}{3} \). Slope of line L (\( m_L \)) must satisfy \( m_1 m_L = -1 \): \[ m_L = -\frac{3}{4} \] The angle \( \alpha \) with the positive X-axis satisfies \( \tan \alpha = m_L \). \[ \alpha = \tan^{-1}\left(-\frac{3}{4}\right) \]
Step 4: Final Answer:
The angle is \( \tan^{-1}\left(-\frac{3}{4}\right) \).