Question

Find the equation of a line with a slope of 13/2 that passes through the point (−2, — 10).

Answer 1

Answer:  $y=\frac{13}{2} x+3$

Step-by-step explanation:

Remember the point-slope equation which is $y-y_{1} = m(x-x_{1} )$  where$(x_{1} , y_{1} )$ is your point and $m$ is your slope.

Given that we substitute what you have:

$y-(-10) =\frac{13}{2} (x - (-2))$

Minus and minus give us positive:

$y+10 =\frac{13}{2} (x +2)$

Multiply slope into the parenthesis:

$y+10= \frac{13}{2}x + \frac{13}{2}*2$

Calculate it:

$y+10= \frac{13}{2}x +13$

Isolate the $y$ by subtracting 10 from both sides:

$y=\frac{13}{2} x + 13 -10$

Your final equation is:

$y=\frac{13}{2} x +3$

Hope this makes sense!

And here's the graph to prove that the line actually goes through the point $(-2,-10)$: