Question

Write the equation of a line that is perpendicular to the given line and that passes through the given point.

–x + 5y = 14; (–5, –2)


y = 5x – 27


y = –5x – 27


y = x – 27


y = x – 15

Answer 1

To find the equation of a line that is perpendicular to the given line \(-x + 5y = 14\) and passes through the point \((-5, -2)\), we first need to determine the slope of the given line.

1. **Rearrange the given line into slope-intercept form (y = mx + b)**:

Starting with:
\[
-x + 5y = 14
\]
Add \(x\) to both sides:
\[
5y = x + 14
\]
Divide by \(5\):
\[
y = \frac{1}{5}x + \frac{14}{5}
\]

The slope of the given line is \(\frac{1}{5}\).

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 slope of that line. Therefore, the slope of the perpendicular line is:
\[
-\frac{1}{\frac{1}{5}} = -5
\]

3. **Use the point-slope form of the equation of a line to find the equation of the perpendicular line**:

The point-slope form is given by:
\[
y - y_1 = m(x - x_1)
\]
where \((x_1, y_1)\) is the point \((-5, -2)\) and \(m = -5\).

Substituting in these values:
\[
y - (-2) = -5(x - (-5))
\]
Simplifying:
\[
y + 2 = -5(x + 5)
\]
\[
y + 2 = -5x - 25
\]
\[
y = -5x - 27
\]

Therefore, the equation of the line that is perpendicular to \(-x + 5y = 14\) and passes through the point \((-5, -2)\) is:
\[
\boxed{y = -5x - 27}
\]