Question

consider the line 6x + 2y = -1. find the equation of the line that is perpendicular to this line and passes through the point (-8, 1). find the equation of the line that is parallel to this line and passes through the point (-8, 1). equation of perpendicular line: equation of parallel line:

Explanation:

Step1: Find slope of given line

Rewrite \(6x + 2y=-1\) in slope - intercept form \(y = mx + b\) (where \(m\) is the slope).
Subtract \(6x\) from both sides: \(2y=-6x - 1\).
Divide by 2: \(y=-3x-\frac{1}{2}\). So the slope of the given line \(m=-3\).

Step2: Find slope of perpendicular line

If two lines are perpendicular, the product of their slopes is \(- 1\). Let the slope of the perpendicular line be \(m_1\).
We have \(m\times m_1=-1\), substitute \(m = - 3\): \(-3\times m_1=-1\).
Solve for \(m_1\): \(m_1=\frac{1}{3}\).

Step3: Equation of perpendicular line

Use point - slope form \(y - y_1=m_1(x - x_1)\) where \((x_1,y_1)=(-8,1)\) and \(m_1=\frac{1}{3}\).
\(y - 1=\frac{1}{3}(x + 8)\).
Expand: \(y-1=\frac{1}{3}x+\frac{8}{3}\).
Add 1 to both sides: \(y=\frac{1}{3}x+\frac{8}{3}+1=\frac{1}{3}x+\frac{8 + 3}{3}=\frac{1}{3}x+\frac{11}{3}\).
Multiply through by 3 to get standard form (or keep in slope - intercept): \(3y=x + 11\) or \(x-3y=-11\), but slope - intercept is \(y=\frac{1}{3}x+\frac{11}{3}\).

Step4: Slope of parallel line

If two lines are parallel, their slopes are equal. So the slope of the parallel line \(m_2=m=-3\).

Step5: Equation of parallel line

Use point - slope form \(y - y_1=m_2(x - x_1)\) with \((x_1,y_1)=(-8,1)\) and \(m_2=-3\).
\(y - 1=-3(x + 8)\).
Expand: \(y - 1=-3x-24\).
Add 1 to both sides: \(y=-3x-24 + 1=-3x-23\).

Answer:

Equation of perpendicular line: \(y=\frac{1}{3}x+\frac{11}{3}\) (or \(x - 3y=-11\))
Equation of parallel line: \(y=-3x-23\)