Question

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

4x – 12y = 2; (10, –1)


y = –3x + 29


y = 3x + 29

Answer 1

To find the equation of a line that is perpendicular to the given line \(4x - 12y = 2\) and passes through the point \((10, -1)\), we need to follow these steps:

1. **Rewrite the given line in slope-intercept form (y = mx + b):**
\[
4x - 12y = 2
\]
First, isolate \(y\):
\[
-12y = -4x + 2
\]
Then, divide by -12:
\[
y = \frac{4}{12}x - \frac{2}{12}
\]
Simplifying gives:
\[
y = \frac{1}{3}x - \frac{1}{6}
\]

The slope \(m_1\) of the original line is \(\frac{1}{3}\).

2. **Find the slope of the perpendicular line:**
The slope of a line perpendicular to another is the negative reciprocal of the original line's slope. Thus,
\[
m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{1}{3}} = -3
\]

3. **Use the point-slope form to write the equation of the perpendicular line:**
The point-slope form of a line is given by:
\[
y - y_1 = m(x - x_1)
\]
Plugging in the point \((10, -1)\) and the new slope \(-3\):
\[
y - (-1) = -3(x - 10)
\]
Simplifying this:
\[
y + 1 = -3(x - 10)
\]
\[
y + 1 = -3x + 30
\]
\[
y = -3x + 30 - 1
\]
\[
y = -3x + 29
\]

Thus, the equation of the line that is perpendicular to \(4x - 12y = 2\) and passes through the point \((10, -1)\) is:
\[
\boxed{y = -3x + 29}
\]