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find the equation of the line parallel to $y = -\frac{9}{4}x - 3$, that…

Question

find the equation of the line parallel to $y = -\frac{9}{4}x - 3$, that passes through $(-8, 4)$.
$y = -\frac{?}{?}x + \square$

Explanation:

Step1: Determine the slope

Parallel lines have the same slope. The given line is \( y = -\frac{9}{4}x - 3 \), so the slope \( m \) of the parallel line is also \( -\frac{9}{4} \).

Step2: Use point - slope form

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(-8,4) \) and \( m =-\frac{9}{4} \).
Substitute the values into the formula:
\( y - 4=-\frac{9}{4}(x + 8) \)

Step3: Simplify the equation

First, distribute \( -\frac{9}{4} \) on the right - hand side:
\( y - 4=-\frac{9}{4}x-18 \)
Then, add 4 to both sides of the equation:
\( y=-\frac{9}{4}x-18 + 4 \)
\( y=-\frac{9}{4}x-14 \)

Answer:

The equation of the line is \( y = -\frac{9}{4}x-14 \), so the first box (numerator of the slope) is \( 9 \), the second box (denominator of the slope) is \( 4 \), and the third box (the constant term) is \( - 14 \).