Question

The slope of a line perpendicular to 13x-7y-1-0 is

• A. \(\frac{13}{7}\)
• B. \(-\frac{13}{7}\)
• C. \(\frac{7}{13}\)
• D. \(-\frac{7}{13}\)

Answer 1

The Correct Option is D

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 \( \mathbf{(4)} \).