Question

The slope of a line L passing through the point \((-2, -3)\) is not defined. If the angle between the lines L and \( ax - 2y + 3 = 0 \) (where \( a>0 \)) is 45°, then the angle made by the line \( x + ay - 4 = 0 \) with positive X-axis in the anticlockwise direction is:

• A. \( \pi - \tan^{-1}\left(\frac{1}{2}\right) \)
• B. \( \frac{\pi}{3} \)
• C. \( \frac{2\pi}{3} \)
• D. \( \tan^{-1}\left(\frac{1}{2}\right) \)

Hint:

For lines with undefined slopes, the perpendicular line will have a slope of zero, making angle calculations with the horizontal straightforward.

Answer 1

The Correct Option is A

Since the slope of line L is not defined, L is a vertical line.
Since it passes through \((-2, -3)\), the equation of L is \(x = -2\).
The slope of the line \(ax - 2y + 3 = 0\) is \(m_1 = \frac{a}{2}\).
The slope of L is undefined, which means the line is vertical.
The angle between L and \(ax - 2y + 3 = 0\) is 45°.
The angle between a vertical line and a line with slope \(m_1\) is given by \(\tan \theta = \frac{1}{|m_1|}\).
\[\tan 45^\circ = \frac{1}{|\frac{a}{2}|}\] \[1 = \frac{2}{a}\] \[a = 2\] Now we need to find the angle made by the line \(x + ay - 4 = 0\) with the positive X-axis.
Substituting \(a = 2\), the equation becomes \(x + 2y - 4 = 0\).
The slope of this line is \(m_2 = -\frac{1}{2}\).
Let \(\alpha\) be the angle made by this line with the positive X-axis.
Then \(\tan \alpha = m_2 = -\frac{1}{2}\).
Since \(\tan \alpha<0\), \(\alpha\) is in the second or fourth quadrant.
Since we want the angle in the anticlockwise direction, we consider the second quadrant.
\[\alpha = \pi - \tan^{-1}\left(\frac{1}{2}\right)\] Final Answer: The final answer is $\boxed{(1)}$