a line has a slope of $-\frac{3}{5}$ and passes through the point $(12, -8)$. write its equation in slope…

Answer:

$y = -\frac{3}{5}x+\frac{16}{5}$

Explanation:

Step1: Recall slope - intercept form

The slope - intercept form of a line is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. We know that $m=-\frac{3}{5}$, so the equation of the line is $y = -\frac{3}{5}x + b$.

Step2: Substitute the point values

We substitute the point $(x = 12,y=-8)$ into the equation $y = -\frac{3}{5}x + b$. So, $-8=-\frac{3}{5}\times12 + b$.

Step3: Calculate the right - hand side

First, calculate $-\frac{3}{5}\times12=-\frac{36}{5}$. Then the equation becomes $-8=-\frac{36}{5}+b$.

Step4: Solve for b

Add $\frac{36}{5}$ to both sides of the equation. $b=-8+\frac{36}{5}$. Rewrite $-8$ as $-\frac{40}{5}$. Then $b = -\frac{40}{5}+\frac{36}{5}=\frac{-40 + 36}{5}=-\frac{4}{5}$.

Step5: Write the final equation

Substitute $b = \frac{16}{5}$ back into $y=mx + b$. The equation of the line in slope - intercept form is $y=-\frac{3}{5}x+\frac{16}{5}$.