Question

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consider the line -7x - 3y = 1.
find the equation of the line that is parallel to this line and passes through the point (7, -4).
find the equation of the line that is perpendicular to this line and passes through the point (7, -4).
equation of parallel line:
equation of perpendicular line:

Explanation:

Step1: Find slope of given line

Rewrite \(-7x - 3y = 1\) in slope - intercept form \(y=mx + b\) (where \(m\) is slope).
Solve for \(y\):
\(-3y=7x + 1\)
\(y=-\frac{7}{3}x-\frac{1}{3}\).
So, slope of given line \(m = -\frac{7}{3}\).

Step2: Equation of parallel line

Parallel lines have equal slopes. So, slope of parallel line \(m_{parallel}=-\frac{7}{3}\).
Use point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(7,-4)\) and \(m = -\frac{7}{3}\):
\(y-(-4)=-\frac{7}{3}(x - 7)\)
\(y + 4=-\frac{7}{3}x+\frac{49}{3}\)
\(y=-\frac{7}{3}x+\frac{49}{3}-4\)
\(y=-\frac{7}{3}x+\frac{49 - 12}{3}\)
\(y=-\frac{7}{3}x+\frac{37}{3}\).
Multiply through by 3 to get standard form: \(7x+3y = 37\).

Step3: Equation of perpendicular line

Perpendicular lines have slopes that are negative reciprocals. So, slope of perpendicular line \(m_{perpendicular}=\frac{3}{7}\) (since \(-\frac{7}{3}\times\frac{3}{7}=-1\)).
Use point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(7,-4)\) and \(m=\frac{3}{7}\):
\(y-(-4)=\frac{3}{7}(x - 7)\)
\(y + 4=\frac{3}{7}x-3\)
\(y=\frac{3}{7}x-3 - 4\)
\(y=\frac{3}{7}x-7\).
Multiply through by 7 to get standard form: \(3x - 7y=49\).

Answer:

Equation of parallel line: \(7x + 3y=37\) (or \(y = -\frac{7}{3}x+\frac{37}{3}\))
Equation of perpendicular line: \(3x - 7y = 49\) (or \(y=\frac{3}{7}x - 7\))