Question

consider the line -4x - 6y = 3. find the equation of the line that is parallel to this line and passes through the point (2, 3). find the equation of the line that is perpendicular to this line and passes through the point (2, 3). 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 \(-4x - 6y = 3\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope).
First, solve for \(y\):
\(-6y=4x + 3\)
\(y=-\frac{4}{6}x-\frac{3}{6}\)
Simplify: \(y =-\frac{2}{3}x-\frac{1}{2}\). So the slope \(m\) of the given line is \(-\frac{2}{3}\).

Step2: Equation of parallel line

Parallel lines have the same slope. So the slope of the parallel line \(m_{parallel}=-\frac{2}{3}\).
Use the point - slope form \(y - y_1=m(x - x_1)\) where \((x_1,y_1)=(2,3)\) and \(m =-\frac{2}{3}\).
\(y - 3=-\frac{2}{3}(x - 2)\)
Expand: \(y-3=-\frac{2}{3}x+\frac{4}{3}\)
Add 3 to both sides: \(y=-\frac{2}{3}x+\frac{4}{3}+3\)
\(y=-\frac{2}{3}x+\frac{4 + 9}{3}\)
\(y=-\frac{2}{3}x+\frac{13}{3}\)
Multiply through by 3 to get standard form: \(3y=-2x + 13\) or \(2x+3y = 13\).

Step3: Equation of perpendicular line

The slope of a perpendicular line \(m_{perpendicular}\) is the negative reciprocal of the original slope. So \(m_{perpendicular}=\frac{3}{2}\) (since the negative reciprocal of \(-\frac{2}{3}\) is \(\frac{3}{2}\)).
Use the point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(2,3)\) and \(m=\frac{3}{2}\).
\(y - 3=\frac{3}{2}(x - 2)\)
Expand: \(y-3=\frac{3}{2}x-3\)
Add 3 to both sides: \(y=\frac{3}{2}x\)
Multiply through by 2: \(2y = 3x\) or \(3x-2y=0\).

Answer:

Equation of parallel line: \(2x + 3y=13\) (or \(y=-\frac{2}{3}x+\frac{13}{3}\))
Equation of perpendicular line: \(3x - 2y = 0\) (or \(y=\frac{3}{2}x\))