To solve this problem, follow these steps:
1. Determine the slope of the original path:
The given equation for the original path is \( y = -3x - 6 \). This equation is in the form \( y = mx + b \), where \( m \) is the slope.
The slope \( m \) of the original path is \( -3 \).
2. Find the slope of the perpendicular path:
For two lines to be perpendicular, the product of their slopes must be \( -1 \). If the slope of the original path is \( -3 \), then let the slope of the perpendicular path be \( m_{\text{perp}} \). We have:
[tex]\[
m \cdot m_{\text{perp}} = -1
\][/tex]
Substituting the slope of the original path:
[tex]\[
-3 \cdot m_{\text{perp}} = -1
\][/tex]
Solving for \( m_{\text{perp}} \):
[tex]\[
m_{\text{perp}} = \frac{1}{-3} = \frac{1}{3}
\][/tex]
So, the slope of the perpendicular path is \(\frac{1}{3}\).
3. Use the point-slope form to write the equation of the new path:
The new path must pass through the point of intersection, \((-3, 3)\). Using the point-slope form of the equation of a line, which is:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
where \( (x_1, y_1) \) is the point of intersection and \( m \) is the slope of the new path, we can plug in the values:
[tex]\[
y - 3 = \frac{1}{3}(x + 3)
\][/tex]
4. Simplify the equation to slope-intercept form \(y = mx + b\):
Distribute \( \frac{1}{3} \) on the right side:
[tex]\[
y - 3 = \frac{1}{3}x + 1
\][/tex]
Add 3 to both sides to isolate \( y \):
[tex]\[
y = \frac{1}{3}x + 4
\][/tex]
So, the equation of the new path is \( y = \frac{1}{3}x + 4 \).
Thus, the correct answer is:
C. [tex]\( y = \frac{1}{3}x + 4 \)[/tex]
