Question

line c has an equation of $y = -\frac{3}{4}x - 3$. line d, which is parallel to line c, includes the point $(-2, -2)$. what is the equation of line d?
write the equation in slope-intercept form. write the numbers in the equation as simplified proper fractions, improper fractions, or integers.

Explanation:

Step1: Determine the slope of line d

Parallel lines have the same slope. The equation of line c is \( y = -\frac{3}{4}x - 3 \), which is in slope - intercept form \( y=mx + b \) (where \( m \) is the slope and \( b \) is the y - intercept). So the slope of line c, \( m=-\frac{3}{4} \), and the slope of line d (since it is parallel to line c) is also \( m = -\frac{3}{4} \).

Step2: Use the point - slope form to find the equation of line d

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1) \) is a point on the line and \( m \) is the slope. We know that \( m = -\frac{3}{4} \) and the point \( (x_1,y_1)=(-2,-2) \). Substitute these values into the point - slope form:
\( y-(-2)=-\frac{3}{4}(x - (-2)) \)
Simplify the left - hand side and the right - hand side:
\( y + 2=-\frac{3}{4}(x + 2) \)

Step3: Convert to slope - intercept form

First, distribute the \( -\frac{3}{4} \) on the right - hand side:
\( y+2=-\frac{3}{4}x-\frac{3}{4}\times2 \)
\( y + 2=-\frac{3}{4}x-\frac{3}{2} \)
Then, subtract 2 from both sides to solve for \( y \). We know that \( 2=\frac{4}{2} \), so:
\( y=-\frac{3}{4}x-\frac{3}{2}-\frac{4}{2} \)
\( y=-\frac{3}{4}x-\frac{7}{2} \)

Answer:

\( y = -\frac{3}{4}x-\frac{7}{2} \)