Question

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

Answer 1

Answer:
`y=(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) =(13)/(2) (x - (-2))`

Minus and minus give us positive:

`y+10 =(13)/(2) (x +2)`

Multiply slope into the parenthesis:

`y+10= (13)/(2)x + (13)/(2)*2\\`

Calculate it:

`y+10= (13)/(2)x +13`

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

`y=(13)/(2) x + 13 -10`

Your final equation is:

`y=(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)`: