Question:
Find the equation of the line perpendicular to the line
\[
7x - 5y = 11
\]
and passing through the point \( (7, -9) \).
Find the equation of the line perpendicular to the line
\[
7x - 5y = 11
\]
and passing through the point \( (7, -9) \).
Show Hint
To find the perpendicular line, take the negative reciprocal of the given line's slope.
Updated On: Apr 7, 2026
- \( 5x + 7y + 28 = 0 \)
- \( 5x + 7y - 28 = 0 \)
- \( 5x + 7y + 38 = 0 \)
- \( 5x + 7y - 38 = 0 \)
- \( 5x - 7y + 28 = 0 \)
Show Solution
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The Correct Option is 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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