What is the equation of a line perpendicular to y = -3x + 4 that passes through the point (-1, 6)?
The correct Answer and Explanation is:
To find the equation of a line perpendicular to ( y = -3x + 4 ) that passes through the point ((-1, 6)), we can follow these steps:
Step 1: Determine the slope of the given line
The equation ( y = -3x + 4 ) is in slope-intercept form ( y = mx + b ), where ( m ) is the slope. The slope of this line is ( m = -3 ).
Step 2: Find the slope of the perpendicular line
Lines that are perpendicular to each other have slopes that are negative reciprocals. To find the negative reciprocal of (-3), we take:
[
m_{\text{perpendicular}} = -\frac{1}{m} = -\frac{1}{-3} = \frac{1}{3}
]
Step 3: Use the point-slope form to find the equation
We will now use the point-slope form of a linear equation, which is given by:
[
y – y_1 = m(x – x_1)
]
Here, ( m ) is the slope we found, ( \frac{1}{3} ), and ((x_1, y_1)) is the point ((-1, 6)).
Substituting the values into the point-slope form:
[
y – 6 = \frac{1}{3}(x – (-1))
]
This simplifies to:
[
y – 6 = \frac{1}{3}(x + 1)
]
Step 4: Rearrange to slope-intercept form
Next, we distribute ( \frac{1}{3} ) on the right side:
[
y – 6 = \frac{1}{3}x + \frac{1}{3}
]
Now, add ( 6 ) to both sides to isolate ( y ):
[
y = \frac{1}{3}x + \frac{1}{3} + 6
]
Convert ( 6 ) into thirds:
[
6 = \frac{18}{3}
]
So, the equation becomes:
[
y = \frac{1}{3}x + \frac{1}{3} + \frac{18}{3} = \frac{1}{3}x + \frac{19}{3}
]
Final Answer
Thus, the equation of the line perpendicular to ( y = -3x + 4 ) and passing through the point ((-1, 6)) is:
[
\boxed{y = \frac{1}{3}x + \frac{19}{3}}
]
This equation describes a line with a slope of (\frac{1}{3}), indicating it rises slowly compared to the steep descent of the original line with a slope of (-3).