Answer:
y = -1/3x + 2
Step-by-step explanation:
Given:
- Original line equation: y = 3x + 2
- Point through which the perpendicular line passes: (6, 0)
To find the equation of a straight line perpendicular to y = 3x + 2 and passing through (6, 0):
1. Determine the slope of the given line:
The slope (m) of y = 3x + 2 is 3.
2. Find the slope of the perpendicular line:
[tex]m_{\text{perpendicular}} = -\frac{1}{3}[/tex]
3. Use the point-slope form of the equation of a line:
[tex]y - y_1 = m(x - x_1)[/tex]
Where:
- m = -1/3
- (x₁, y₁) = (6, 0)
4. Substitute the values into the point-slope form:
[tex]y - 0 = -\frac{1}{3}(x - 6)[/tex]
Simplify:
[tex]y = -\frac{1}{3}(x - 6)[/tex]
[tex]y = -\frac{1}{3}x + 2[/tex]
Therefore, the equation of the line perpendicular to y = 3x + 2 and passing through (6, 0) is:
[tex] \boxed{y = -\frac{1}{3}x + 2}[/tex]
