A walking path across a park is represented by the equation [tex]y = -3x - 6[/tex]. A new path will be built perpendicular to this path. The paths will intersect at the point [tex](-3, 3)[/tex]. Identify the equation that represents the new path.
A. [tex]y = 3x + 12[/tex] B. [tex]y = -3x - 6[/tex] C. [tex]y = \frac{1}{3}x + 4[/tex] D. [tex]y = -\frac{1}{3}x + 2[/tex]
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]
