Question

consider the line 8x - 3y = -6.
find the equation of the line that is parallel to this line and passes through the point (-6, 4).
find the equation of the line that is perpendicular to this line and passes through the point (-6, 4).
note that the aleks graphing calculator may be helpful in checking your answer.
equation of parallel line:
equation of perpendicular line:

Explanation:

Step1: Find slope of given line

Rewrite \(8x - 3y=-6\) in slope - intercept form \(y = mx + b\) (where \(m\) is the slope).
\(-3y=-8x - 6\)
Divide both sides by \(-3\): \(y=\frac{8}{3}x + 2\). So the slope \(m\) of the given line is \(\frac{8}{3}\).

Step2: Equation of parallel line

Parallel lines have the same slope. So the slope of the parallel line \(m_{parallel}=\frac{8}{3}\).
Use point - slope form \(y - y_1=m(x - x_1)\) where \((x_1,y_1)=(-6,4)\) and \(m = \frac{8}{3}\).
\(y - 4=\frac{8}{3}(x+6)\)
Expand: \(y - 4=\frac{8}{3}x+16\)
Add 4 to both sides: \(y=\frac{8}{3}x + 20\). Multiply through by 3 to get standard form: \(8x-3y=-60\) (or keep in slope - intercept form \(y=\frac{8}{3}x + 20\)).

Step3: Equation of perpendicular line

The slope of a perpendicular line \(m_{perpendicular}=-\frac{1}{m}\) (negative reciprocal). Since \(m = \frac{8}{3}\), \(m_{perpendicular}=-\frac{3}{8}\).
Use point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(-6,4)\) and \(m=-\frac{3}{8}\).
\(y - 4=-\frac{3}{8}(x + 6)\)
Expand: \(y - 4=-\frac{3}{8}x-\frac{18}{8}\)
Simplify \(-\frac{18}{8}=-\frac{9}{4}\), then \(y=-\frac{3}{8}x-\frac{9}{4}+4\)
\(4=\frac{16}{4}\), so \(y=-\frac{3}{8}x+\frac{16 - 9}{4}=-\frac{3}{8}x+\frac{7}{4}\). Multiply through by 8 to get standard form: \(3x + 8y = 14\) (or keep in slope - intercept form \(y=-\frac{3}{8}x+\frac{7}{4}\)).

Answer:

Equation of parallel line: \(y=\frac{8}{3}x + 20\) (or \(8x-3y=-60\))
Equation of perpendicular line: \(y=-\frac{3}{8}x+\frac{7}{4}\) (or \(3x + 8y = 14\))