Question

consider the line $-6x - 3y = 8$. find the equation of the line that is perpendicular to this line and passes through the point $(4, 3)$. find the equation of the line that is parallel to this line and passes through the point $(4, 3)$. 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 \(-6x - 3y = 8\) in slope - intercept form \(y=mx + b\) (where \(m\) is slope).
Add \(6x\) to both sides: \(-3y=6x + 8\).
Divide by \(-3\): \(y=-2x-\frac{8}{3}\).
So slope of given line \(m=-2\).

Step2: Equation of perpendicular line

Slope of perpendicular line \(m_{\perp}\) satisfies \(m\times m_{\perp}=-1\).
Since \(m = - 2\), then \(-2\times m_{\perp}=-1\Rightarrow m_{\perp}=\frac{1}{2}\).
Use point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(4,3)\) and \(m=\frac{1}{2}\).
\(y - 3=\frac{1}{2}(x - 4)\).
Simplify: \(y-3=\frac{1}{2}x - 2\Rightarrow y=\frac{1}{2}x+1\).

Step3: Equation of parallel line

Slope of parallel line \(m_{\parallel}=m=-2\) (parallel lines have equal slopes).
Use point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(4,3)\) and \(m = - 2\).
\(y - 3=-2(x - 4)\).
Simplify: \(y-3=-2x + 8\Rightarrow y=-2x+11\).

Answer:

Equation of perpendicular line: \(y=\frac{1}{2}x + 1\)
Equation of parallel line: \(y=-2x + 11\)