Question

Passing through left parenthesis 1 comma negative 1 right parenthesis and perpendicular to the line whose equation is x minus 8 y minus 7 equals 0

find equation

Answer 1

To find the equation of the line that passes through the point \( (1, -1) \) and is perpendicular to the line given by the equation \( x - 8y - 7 = 0 \), we need to follow these steps:

1. **Find the slope of the given line**:
First, we can rewrite the equation \( x - 8y - 7 = 0 \) in slope-intercept form \( y = mx + b \).

\[
x - 8y - 7 = 0 \implies -8y = -x + 7 \implies y = \frac{1}{8}x + \frac{7}{8}
\]

The slope (\( m \)) of the given line is \( \frac{1}{8} \).

2. **Find the slope of the perpendicular line**:
The slope of a line that is perpendicular to another line is the negative reciprocal of the original line's slope. Therefore, the slope \( m_{\text{perpendicular}} \) is:

\[
m_{\text{perpendicular}} = -\frac{1}{\frac{1}{8}} = -8
\]

3. **Use point-slope form to write the equation of the perpendicular line**:
The point-slope form of a line's equation is given by:

\[
y - y_1 = m(x - x_1)
\]

where \( (x_1, y_1) \) is the point through which the line passes, and \( m \) is the slope. Here, \( (x_1, y_1) = (1, -1) \) and \( m = -8 \).

Plugging in these values gives:

\[
y - (-1) = -8(x - 1)
\]

Simplifying this:

\[
y + 1 = -8x + 8
\]

\[
y = -8x + 8 - 1
\]

\[
y = -8x + 7
\]

4. **Final equation**:
The equation of the line that passes through the point \( (1, -1) \) and is perpendicular to the line \( x - 8y - 7 = 0 \) is:

\[
y = -8x + 7
\]

Alternatively, you can express it in standard form. Rearranging gives:

\[
8x + y - 7 = 0
\]

Thus, either \( y = -8x + 7 \) or \( 8x + y - 7 = 0 \) are acceptable forms for the equation of the required line.