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 = x + 7


y = 3x + 29


y = x + 29


y = –3x + 29

Answer 1

To find the equation of the line that is perpendicular to the given line \(4x - 12y = 2\) and passes through the point \((10, -1)\), we first need to determine the slope of the given line.

1. **Convert the given line to slope-intercept form (y = mx + b)**:
\[
4x - 12y = 2
\]
Rearranging gives:
\[
-12y = -4x + 2
\]
Dividing by -12:
\[
y = \frac{1}{3}x - \frac{1}{6}
\]
The slope (m) of the given line is \(\frac{1}{3}\).

2. **Find the slope of the perpendicular line**:
The slope of a line that is perpendicular to another is the negative reciprocal of the original slope. Therefore:
\[
m_{\text{perpendicular}} = -\frac{1}{(\frac{1}{3})} = -3
\]

3. **Use the point-slope form to write the equation of the perpendicular line that passes through (10, -1)**:
Point-slope form is given by:
\[
y - y_1 = m(x - x_1)
\]
Substituting \(m = -3\), \(x_1 = 10\), and \(y_1 = -1\):
\[
y - (-1) = -3(x - 10)
\]
Simplifying:
\[
y + 1 = -3x + 30
\]
\[
y = -3x + 29
\]

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