Question

consider the line $y = 7x - 6$. find the equation of the line that is parallel to this line and passes through the point $(5, -2)$. find the equation of the line that is perpendicular to this line and passes through the point $(5, -2)$. equation of parallel line: equation of perpendicular line:

Explanation:

Step1: Recall slope of parallel lines

Parallel lines have equal slopes. The given line is \( y = 7x - 6 \), so its slope \( m = 7 \). The parallel line will also have slope \( 7 \).
Use point - slope form \( y - y_1 = m(x - x_1) \), where \( (x_1,y_1)=(5,-2) \) and \( m = 7 \).
\( y - (-2)=7(x - 5) \)
\( y + 2 = 7x-35 \)
\( y=7x - 35 - 2 \)
\( y = 7x-37 \)

Step2: Recall slope of perpendicular lines

Perpendicular lines have slopes that are negative reciprocals. If the slope of the given line is \( m = 7 \), the slope of the perpendicular line \( m_{\perp}=-\frac{1}{7} \).
Use point - slope form \( y - y_1=m_{\perp}(x - x_1) \), with \( (x_1,y_1)=(5,-2) \) and \( m_{\perp}=-\frac{1}{7} \).
\( y-(-2)=-\frac{1}{7}(x - 5) \)
\( y + 2=-\frac{1}{7}x+\frac{5}{7} \)
\( y=-\frac{1}{7}x+\frac{5}{7}-2 \)
\( y=-\frac{1}{7}x+\frac{5 - 14}{7} \)
\( y=-\frac{1}{7}x-\frac{9}{7} \)

Answer:

Equation of parallel line: \( y = 7x - 37 \)
Equation of perpendicular line: \( y=-\frac{1}{7}x-\frac{9}{7} \)