Finding a Parallel Line Passing Through a Point
The equation for line q is given as:
\( y = -5 - \frac{1}{8}(x + 2) \)
To find a line parallel to q that passes through the point (-6, 1), we need to keep the slope the same, since parallel lines have equal slopes.
First, let's rewrite the equation for line q in slope-intercept form (\( y = mx + b \)), where \( m \) is the slope and \( b \) is the y-intercept.
Rewrite the equation of line q to make the slope more apparent:
\( y = -5 - \frac{1}{8}x - \frac{1}{8}(2) \)
\( y = -5 - \frac{1}{8}x - \frac{1}{4} \)
\( y = -\frac{1}{8}x - 5 - \frac{1}{4} \)
\( y = -\frac{1}{8}x - 5.25 \)
\( y = -\frac{1}{8}x - \frac{21}{4} \)
Now we know the slope of line q is \( -\frac{1}{8} \).
Since line r is parallel to line q, it will also have a slope of \( -\frac{1}{8} \).
Using the point-slope form of a line's equation ( \( y - y_1 = m(x - x_1) \) ), where \( m \) is the slope and \( (x_1, y_1) \) is the point (-6, 1) through which line r passes, we can write:
\( y - 1 = -\frac{1}{8}(x - (-6)) \)
\( y - 1 = -\frac{1}{8}(x + 6) \)
Now we can put this in slope-intercept form:
\( y = -\frac{1}{8}x - \frac{1}{8}(6) + 1 \)
\( y = -\frac{1}{8}x - \frac{3}{4} + 1 \)
\( y = -\frac{1}{8}x + \frac{1}{4} \)
So, the equation of line r in slope-intercept form is:
\( y = -\frac{1}{8}x + \frac{1}{4} \)
