Question

The equation of a straight line passing through the point of intersection of \( x - y + 1 = 0 \) and \( 3x + y - 5 = 0 \) and perpendicular to one of them, is:

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

Hint:

For a line perpendicular to a given line, the product of the slopes of both lines will be \( -1 \).

Answer 1

The Correct Option is D

The equations of the lines are: \[ x - y + 1 = 0 \quad \text{(1)} \] \[ 3x + y - 5 = 0 \quad \text{(2)} \] To find the point of intersection of these lines, we solve the system of equations (1) and (2): Adding equations (1) and (2) gives: \[ x - y + 1 + 3x + y - 5 = 0 \] \[ 4x - 4 = 0 \] \[ x = 1 \] Substitute \( x = 1 \) in equation (1) to find \( y \): \[ 1 - y + 1 = 0 \quad \Rightarrow \quad y = 2 \] So, the point of intersection is \( (1, 2) \). Now, the slope of line (1) is \( m_1 = 1 \) and the slope of line (2) is \( m_2 = -3 \).
The slope of the perpendicular line is the negative reciprocal of \( m_2 \): \[ m = \frac{1}{3} \] Using the point-slope form, the equation of the line passing through \( (1, 2) \) with slope \( \frac{1}{3} \) is: \[ y - 2 = \frac{1}{3}(x - 1) \] Simplifying: \[ y - 2 = \frac{1}{3}x - \frac{1}{3} \quad \Rightarrow \quad 3y - 6 = x - 1 \] \[ x - 3y + 5 = 0 \] Thus, the correct equation is \( x - 3y + 5 = 0 \).