Question

consider the line y=9x−3. find the equation of the line that is parallel to this line and passes through the point (−6, −3). find the equation of the line that is perpendicular to this line and passes through the point (−6, −3). note that the aleks graphing calculator may be helpful in checking your answer. equation of parallel line: equation of perpendicular line:

Explanation:

Step1: Recall slope of parallel lines

Parallel lines have the same slope. The given line is \( y = 9x - 3 \), so its slope \( m = 9 \). The parallel line will also have slope \( 9 \).

Step2: Use point - slope form for parallel line

Point - slope form is \( y - y_1 = m(x - x_1) \), where \( (x_1,y_1)=(-6,-3) \) and \( m = 9 \).
Substitute values: \( y - (-3)=9(x - (-6)) \)
Simplify: \( y + 3 = 9(x + 6) \)
Expand: \( y+3 = 9x+54 \)
Subtract 3: \( y = 9x + 51 \)

Step3: Recall slope of perpendicular lines

Perpendicular lines have slopes that are negative reciprocals. If \( m_1 = 9 \), then \( m_2=-\frac{1}{9} \).

Step4: Use point - slope form for perpendicular line

Using point - slope form with \( (x_1,y_1)=(-6,-3) \) and \( m =-\frac{1}{9} \).
\( y - (-3)=-\frac{1}{9}(x - (-6)) \)
Simplify: \( y + 3=-\frac{1}{9}(x + 6) \)
Expand: \( y+3=-\frac{1}{9}x-\frac{6}{9} \)
Simplify \( \frac{6}{9}=\frac{2}{3} \): \( y + 3=-\frac{1}{9}x-\frac{2}{3} \)
Subtract 3: \( y=-\frac{1}{9}x-\frac{2}{3}-3 \)
Convert 3 to thirds: \( 3=\frac{9}{3} \), so \( y =-\frac{1}{9}x-\frac{2 + 9}{3}=-\frac{1}{9}x-\frac{11}{3} \)

Answer:

Equation of parallel line: \( y = 9x + 51 \)
Equation of perpendicular line: \( y=-\frac{1}{9}x-\frac{11}{3} \)