Question:
The equation of a line perpendicular to $3x - 4y = 7$ and passing through $(2,5)$ is:
The equation of a line perpendicular to $3x - 4y = 7$ and passing through $(2,5)$ is:
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For perpendicular lines: $m_1 \times m_2 = -1$.
Updated On: Apr 23, 2026
- $4y + 3x = 23$
- $3y + 4x = 23$
- $2x + 5y = 6$
- $2x + 5y = 29$
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The Correct Option is B
Solution and Explanation
Concept:
Slope of perpendicular lines are negative reciprocals.
Step 1: Find slope of given line.
\[ 3x - 4y = 7 \Rightarrow y = \frac{3}{4}x - \frac{7}{4} \] \[ m_1 = \frac{3}{4} \]
Step 2: Slope of perpendicular line.
\[ m_2 = -\frac{4}{3} \]
Step 3: Use point-slope form.
\[ y - 5 = -\frac{4}{3}(x - 2) \]
Step 4: Simplify.
\[ 3y - 15 = -4x + 8 \] \[ 4x + 3y = 23 \]
Hence, the equation is $3y + 4x = 23$.
Step 1: Find slope of given line.
\[ 3x - 4y = 7 \Rightarrow y = \frac{3}{4}x - \frac{7}{4} \] \[ m_1 = \frac{3}{4} \]
Step 2: Slope of perpendicular line.
\[ m_2 = -\frac{4}{3} \]
Step 3: Use point-slope form.
\[ y - 5 = -\frac{4}{3}(x - 2) \]
Step 4: Simplify.
\[ 3y - 15 = -4x + 8 \] \[ 4x + 3y = 23 \]
Hence, the equation is $3y + 4x = 23$.
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