Question

consider the line $x + 7y = -6$. find the equation of the line that is perpendicular to this line and passes through the point $(6, -5)$. find the equation of the line that is parallel to this line and passes through the point $(6, -5)$. 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 the slope of the given line

First, rewrite the given line \(x + 7y=-6\) in slope - intercept form \(y = mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept).
Solve for \(y\):
\(7y=-x - 6\)
\(y=-\frac{1}{7}x-\frac{6}{7}\)
So the slope of the given line \(m_1 =-\frac{1}{7}\).

Step2: Find the slope of the perpendicular line

If two lines are perpendicular, the product of their slopes is \(- 1\). Let the slope of the perpendicular line be \(m_2\). Then \(m_1\times m_2=-1\).
Substitute \(m_1 =-\frac{1}{7}\) into the formula:
\(-\frac{1}{7}\times m_2=-1\)
Multiply both sides by \(-7\): \(m_2 = 7\)

Step3: Find the equation of the perpendicular line

We use the point - slope form of a line \(y - y_1=m(x - x_1)\), where \((x_1,y_1)=(6,-5)\) and \(m = 7\).
Substitute the values:
\(y-(-5)=7(x - 6)\)
\(y + 5=7x-42\)
\(y=7x-42 - 5\)
\(y=7x-47\)
We can also write it in standard form: \(7x-y = 47\)

Step4: Find the slope of the parallel line

If two lines are parallel, their slopes are equal. So the slope of the parallel line \(m_3=m_1=-\frac{1}{7}\)

Step5: Find the equation of the parallel line

Use the point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(6,-5)\) and \(m =-\frac{1}{7}\)
\(y-(-5)=-\frac{1}{7}(x - 6)\)
\(y + 5=-\frac{1}{7}x+\frac{6}{7}\)
\(y=-\frac{1}{7}x+\frac{6}{7}-5\)
\(y=-\frac{1}{7}x+\frac{6 - 35}{7}\)
\(y=-\frac{1}{7}x-\frac{29}{7}\)
Multiply through by 7 to get standard form: \(x + 7y=-29\)

Answer:

Equation of perpendicular line: \(y = 7x-47\) (or \(7x - y=47\))
Equation of parallel line: \(y=-\frac{1}{7}x-\frac{29}{7}\) (or \(x + 7y=-29\))