Question

Find the equation of a line passing through(-3,5) and perpendicular through(2,5) & (-3,6)

Answer 1

To find the equation of a line passing through the point \((-3, 5)\) and perpendicular to the line segment between the points \((2, 5)\) and \((-3, 6)\), we first need to determine the slope of that line segment.

1. **Calculate the slope of the line segment between (2, 5) and (-3, 6)**:
The formula for the slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
Here, \((x_1, y_1) = (2, 5)\) and \((x_2, y_2) = (-3, 6)\):
\[
m = \frac{6 - 5}{-3 - 2} = \frac{1}{-5} = -\frac{1}{5}
\]

2. **Determine the slope of the line perpendicular to this line segment**:
The slope of a line perpendicular to another line is the negative reciprocal of the original slope. Thus, if the original slope \( m = -\frac{1}{5} \), then the perpendicular slope \( m_{\perp} \) is:
\[
m_{\perp} = -\frac{1}{-\frac{1}{5}} = 5
\]

3. **Use the point-slope form**:
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:
\[
y - y_1 = m(x - x_1)
\]
Substituting \((x_1, y_1) = (-3, 5)\) and \( m_{\perp} = 5\):
\[
y - 5 = 5(x + 3)
\]

4. **Simplify to get the line equation**:
Now, expand and simplify the equation:
\[
y - 5 = 5x + 15
\]
\[
y = 5x + 15 + 5
\]
\[
y = 5x + 20
\]

Thus, the equation of the line passing through \((-3, 5)\) and perpendicular to the line segment between \((2, 5)\) and \((-3, 6)\) is:
\[
\boxed{y = 5x + 20}
\]