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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]


Sagot :

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]
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