Question

consider the line $x - 5y = -1$. find the equation of the line that is perpendicular to this line and passes through the point $(-5, 5)$. find the equation of the line that is parallel to this line and passes through the point $(-5, 5)$. note that the aleks graphing calculator may be helpful in checking your answer. equation of perpendicular line: $\square$ equation of parallel line: $\square$

Explanation:

Step1: Find slope of given line

Rewrite \(x - 5y=-1\) in slope - intercept form \(y = mx + b\) (where \(m\) is the slope).
Solve for \(y\):
\(-5y=-x - 1\)
\(y=\frac{1}{5}x+\frac{1}{5}\)
So the slope of the given line \(m_1=\frac{1}{5}\).

Step2: Equation of perpendicular line

The slope of a line perpendicular to a line with slope \(m\) is \(m_{\perp}=-\frac{1}{m}\).
For \(m_1 = \frac{1}{5}\), \(m_{\perp}=- 5\).
Use point - slope form \(y - y_1=m(x - x_1)\) where \((x_1,y_1)=(-5,5)\) and \(m = - 5\).
\(y - 5=-5(x + 5)\)
\(y-5=-5x-25\)
\(y=-5x - 20\) or \(5x+y=-20\)

Step3: Equation of parallel line

The slope of a line parallel to a line with slope \(m\) is \(m_{\parallel}=m\).
So \(m_{\parallel}=\frac{1}{5}\).
Use point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(-5,5)\) and \(m=\frac{1}{5}\).
\(y - 5=\frac{1}{5}(x + 5)\)
\(y-5=\frac{1}{5}x + 1\)
\(y=\frac{1}{5}x+6\) or \(x - 5y=-30\)

Answer:

Equation of perpendicular line: \(y=-5x - 20\) (or \(5x + y=-20\))
Equation of parallel line: \(y=\frac{1}{5}x + 6\) (or \(x-5y=-30\))