Question

A line \( L : x + y = 0 \) is given. Two lines \( L_1 \) & \( L_2 \) are passing through (-1, -1) inclined at an angle of 45° from line L. Reflection of lines \( L_1 \) and \( L_2 \) in line \( 2y + x = 1 \) is \( ax + by = 9 \) and \( cx + dy = 1 \) then the value of \( |ad + bc| \) is equal to:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

If a line is parallel or perpendicular to the axes, its reflection is often easier to find by reflecting two points on the line and joining them.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The slope of line \(L\) is \(m = -1\).
If a line is inclined at \(45^\circ\) to a line with slope \(m\), its slope \(m'\) is given by \(\tan 45^\circ = \left| \frac{m' - m}{1 + mm'} \right|\).
Step 2: Key Formula or Approach:
\[ 1 = \left| \frac{m' - (-1)}{1 + (-1)m'} \right| \Rightarrow 1 = \left| \frac{m' + 1}{1 - m'} \right| \]
Case 1: \(1 = \frac{m' + 1}{1 - m'} \Rightarrow 1 - m' = m' + 1 \Rightarrow 2m' = 0 \Rightarrow m' = 0\).
Case 2: \(-1 = \frac{m' + 1}{1 - m'} \Rightarrow m' - 1 = m' + 1 \Rightarrow \text{No solution (line is vertical, slope is undefined/infinite)}\).
So \(L_1\) has slope 0 and \(L_2\) is vertical.
Since they pass through \((-1, -1)\):
\(L_1: y = -1\) and \(L_2: x = -1\).
Step 3: Detailed Explanation:
Now, we find reflections of \(L_1\) and \(L_2\) in the line \(x + 2y = 1\).
For \(L_2: x = -1\) (constant \(x\)):
Take a point on \(x = -1\), say \(P(-1, 1)\). Reflection \(A'\) of \(P\) in \(x + 2y - 1 = 0\):
\[ \frac{x' + 1}{1} = \frac{y' - 1}{2} = \frac{-2(-1 + 2(1) - 1)}{1^2 + 2^2} = 0 \Rightarrow A' = (-1, 1) \]
Since the intersection point of \(x = -1\) and \(x + 2y = 1\) is \((-1, 1)\), the line passes through this point.
Take another point \(Q(-1, -1)\). Its reflection \(A''\):
\[ \frac{x' + 1}{1} = \frac{y' + 1}{2} = \frac{-2(-1 + 2(-1) - 1)}{5} = \frac{8}{5} \]
\[ x' = \frac{3}{5}, \quad y' = \frac{11}{5} \]
The line passing through \((-1, 1)\) and \((3/5, 11/5)\) is \(3x - 4y + 7 = 0\).
Comparing with \(cx + dy = 1\): \(\frac{3}{-7}x + \frac{4}{7}y = 1 \Rightarrow c = -3/7, d = 4/7\).
Similarly, reflecting \(y = -1\) gives the line \(4x + 3y = 9\).
Comparing with \(ax + by = 9\): \(a = 4, b = 3\).
Calculating \(|ad + bc|\):
\[ \left|4 \cdot \frac{4}{7} + 3 \cdot \left(-\frac{3}{7}\right)\right| = \left|\frac{16}{7} - \frac{9}{7}\right| = \left|\frac{7}{7}\right| = 1 \]
Step 4: Final Answer:
The value of \(|ad + bc|\) is 1.