Question

How do you write the equation of a line in slope intercept, point slope and standard form given (-3, -1) and (6, -4)?

Answer 1

The slope intercept form is `y=-1/3x-2`.

Explanation:

First determine the slope, `m`, using the equation:

`m=(y_2-y_1)/(x_2-x_1)`,

where `(x_1,y_1)` is one point and `(x_2,y_2)` is the other point. Either of the given points can be the first or second point. I'm going to use the first point as point 1, and the second point as point 2.

`m=(-4-(-1))/(6-(-3))`

Simplify.

`m=(-4+1)/(6+3)`

`m=-3/9=-1/3`

The slope intercept form is:

`y=mx+b`,

where `b` is the y-intercept. We have slope, and `x` and `y` from one point. Insert the given values into the equation and solve for `b`. I'm going to use the first point `(-3,-1)`.

`-1=-1/3xx-3+b`

Simplify.

`-1=-(-3)/3+b`

`-1=1+b`

Subtract `1` from both sides.

`b=-2`

Slope intercept form is `y=-1/3x-2`.
graph{y=-1/3x-2 [-16.02, 16.02, -8.01, 8.01]}