The slope of a line perpendicular to 13x-7y-1-0 is
- \(\frac{13}{7}\)
- \(-\frac{13}{7}\)
- \(\frac{7}{13}\)
- \(-\frac{7}{13}\)
The Correct Option is D
Solution and Explanation
Step 1: Recall the relationship between slopes of perpendicular lines.
If the slope of a line is \( m \), then the slope of a line perpendicular to it is \( -\frac{1}{m} \).
Step 2: Find the slope of the given line.
The general form of a straight line is \( Ax + By + C = 0 \). The slope of the line is given by:
\[ m = -\frac{\text{coefficient of } x}{\text{coefficient of } y} = -\frac{A}{B}. \]
For the line \( 13x - 7y + 1 = 0 \), \( A = 13 \) and \( B = -7 \). Thus, the slope of the line is:
\[ m = -\frac{13}{-7} = \frac{13}{7}. \]
Step 3: Find the slope of the perpendicular line.
The slope of a line perpendicular to this one is the negative reciprocal of \( \frac{13}{7} \):
\[ \text{Perpendicular slope} = -\frac{1}{\frac{13}{7}} = -\frac{7}{13}. \]
Final Answer: The slope of the line perpendicular to \( 13x - 7y + 1 = 0 \) is \( \mathbf{-\frac{7}{13}} \), which corresponds to option \( \mathbf{(4)} \).
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