Question - Finding a Parallel Line Passing Through a Point
Solution:
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:$$ y = -\frac{1}{8}x + \frac{1}{4} $$
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