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find the equation of the line perpendicular to the
Question:
Find the equation of the line perpendicular to the line \[ 7x - 5y = 11 \] and passing through the point \( (7, -9) \).
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To find the perpendicular line, take the negative reciprocal of the given line's slope.
KEAM - 2024
KEAM
Updated On:
Mar 6, 2025
\( 5x + 7y + 28 = 0 \)
\( 5x + 7y - 28 = 0 \)
\( 5x + 7y + 38 = 0 \)
\( 5x + 7y - 38 = 0 \)
\( 5x - 7y + 28 = 0 \)
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A
Solution and Explanation
The given line equation is: \[ 7x - 5y = 11 \] Slope of the given line: \[ m_1 = \frac{7}{5} \] Since perpendicular slopes are negative reciprocals: \[ m_2 = -\frac{5}{7} \] Using the point-slope form: \[ y - y_1 = m(x - x_1) \] Substituting \( (7, -9) \): \[ y + 9 = -\frac{5}{7} (x - 7) \] Multiplying both sides by 7: \[ 7(y + 9) = -5(x - 7) \] \[ 7y + 63 = -5x + 35 \] \[ 5x + 7y + 28 = 0 \]
Final Answer:
\[ \boxed{5x + 7y + 28 = 0} \]
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