Question

The equation to the line through the point of intersection of \(x - y + 1 = 0\), \(3x + 2y + 4 = 0\) and perpendicular to \(x - 4y = 0\) is

• A. \(4x + y + 5 = 0\)
• B. \(4x + y + 3 = 0\)
• C. \(4x + y - 5 = 0\)
• D. \(4x + y - 3 = 0\)

Hint:

Always solve intersection carefully—small arithmetic errors change the final answer.

Answer 1

The Correct Option is A

Concept: Perpendicular slope = negative reciprocal.

Step 1: Find intersection point.
\[ x - y + 1 = 0 \Rightarrow x = y - 1 \] Substitute into second equation: \[ 3(y-1) + 2y + 4 = 0 \Rightarrow 3y - 3 + 2y + 4 = 0 \Rightarrow 5y + 1 = 0 \Rightarrow y = -\frac{1}{5} \] \[ x = y - 1 = -\frac{1}{5} - 1 = -\frac{6}{5} \]

Step 2: Find slope.
\[ x - 4y = 0 \Rightarrow y = \frac{x}{4} \Rightarrow m = \frac{1}{4} \] \[ m_{\perp} = -4 \]

Step 3: Equation of line.
\[ y + \frac{1}{5} = -4\left(x + \frac{6}{5}\right) \] \[ 5y + 1 = -20x - 24 \Rightarrow 20x + 5y + 25 = 0 \] \[ 4x + y + 5 = 0 \]