Question:
The equation of the line passing through \( (-3, 5) \) and perpendicular to the line through the points \( (1, 0) \) and \( (-4, 1) \) is:
The equation of the line passing through \( (-3, 5) \) and perpendicular to the line through the points \( (1, 0) \) and \( (-4, 1) \) is:
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A line perpendicular to \( Ax + By + C = 0 \) always takes the form \( Bx - Ay + K = 0 \).
Updated On: May 1, 2026
- \( 5x + y + 10 = 0 \)
- \( 5x - y + 20 = 0 \)
- \( 5x - y - 10 = 0 \)
- \( 5x + y + 20 = 0 \)
- \( 5y - x - 10 = 0 \)
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The Correct Option is B
Solution and Explanation
Concept: For two lines to be perpendicular, the product of their slopes must be \( -1 \) (\( m_1 \cdot m_2 = -1 \)). Once the perpendicular slope is found, we use the point-slope form \( y - y_1 = m(x - x_1) \).
Step 1: Calculating the initial and perpendicular slopes.
Slope of the line through \( (1, 0) \) and \( (-4, 1) \): \[ m_1 = \frac{1 - 0}{-4 - 1} = -\frac{1}{5} \] The perpendicular slope \( m_2 \) is the negative reciprocal: \[ m_2 = 5 \]
Step 2: Deriving the final equation.
Using the point \( (-3, 5) \) and slope \( 5 \): \[ y - 5 = 5(x - (-3)) \implies y - 5 = 5x + 15 \] \[ 5x - y + 20 = 0 \]
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