Question

equation for a line perpendicular to y = 3x + 5 that passes through the point (-30, 12)

Answer 1

To find the equation of a line that is perpendicular to the line given by the equation \( y = 3x + 5 \) and passes through the point \((-30, 12)\), we first need to determine the slope of the original line.

The slope of the line \( y = 3x + 5 \) is \( 3 \). The slope of a line that is perpendicular to another line is the negative reciprocal of the original slope. Therefore, the slope \( m \) of the line we want to find is:

\[
m = -\frac{1}{3}
\]

Now that we have the slope of the perpendicular line, we can use the point-slope form of the equation of a line, which is given by:

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

Where \( (x_1, y_1) \) is the point through which the line passes. In this case, the point is \((-30, 12)\):
- \( x_1 = -30 \)
- \( y_1 = 12 \)
- \( m = -\frac{1}{3} \)

Now we can substitute these values into the point-slope formula:

\[
y - 12 = -\frac{1}{3}(x + 30)
\]

Next, we can simplify this equation. First, distribute the slope on the right side:

\[
y - 12 = -\frac{1}{3}x - 10
\]

Now, add \( 12 \) to both sides to isolate \( y \):

\[
y = -\frac{1}{3}x - 10 + 12
\]

Combine like terms:

\[
y = -\frac{1}{3}x + 2
\]

Thus, the equation of the line that is perpendicular to \( y = 3x + 5 \) and passes through the point \((-30, 12)\) is:

\[
y = -\frac{1}{3}x + 2
\]