Questions: Consider the line y=-5x+7. Find the equation of the line that is parallel to this line and passes through the point (-7,5). Find the equation of the line that is perpendicular to this line and passes through the point (-7,5).

Solution Steps

The given line is $y = -5x + 7$. The slope of this line is $m = -5$.

A line parallel to the given line will have the same slope, $m = -5$. Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, and the point $(-7, 5)$, we substitute the values: $$y - 5 = -5(x - (-7))$$ Simplify the equation: $$y - 5 = -5(x + 7)$$ $$y - 5 = -5x - 35$$ $$y = -5x - 30$$

A line perpendicular to the given line will have a slope that is the negative reciprocal of $m = -5$. Thus, the slope of the perpendicular line is: $$m_{\text{perpendicular}} = \frac{1}{5}$$

Using the point-slope form again with the slope $m = \frac{1}{5}$ and the point $(-7, 5)$, we substitute the values: $$y - 5 = \frac{1}{5}(x - (-7))$$ Simplify the equation: $$y - 5 = \frac{1}{5}(x + 7)$$ $$y - 5 = \frac{1}{5}x + \frac{7}{5}$$ $$y = \frac{1}{5}x + \frac{7}{5} + 5$$ $$y = \frac{1}{5}x + \frac{7}{5} + \frac{25}{5}$$ $$y = \frac{1}{5}x + \frac{32}{5}$$

Final Answer