Question

How do you write an equation for the line with slope -2/3 that passes through the point (0,4)?

Answer 1

Refer to the explanation.

Explanation:

Use the point-slope form for a linear equation. The general equation is `y-y_1=m(x-x_1)`, where `x_1` and `y_1` are the known point, and,`m` is the slope.

`"Known point"=(0,4)`
`x_1=0`
`y_1=4`
`m=-2/3`

Substitute the known values and solve for `y`.

`y-y_1=m(x-x_1)`

`y-4=-2/3(x-0)=`

Add `4` to both sides.

`y=-2/3(x-0)+4`

Distribute `-2/3`.

`y=-2/3*x-(-2/3*0)+4`

Simplify.

`y=-2/3x+0+4=`

`y=-2/3x+4`

The equation is now in point-intercept form `y=mx+b`, where `m` is the slope and `b` is the y-intercept.

graph{y=-2/3x+4 [-10.15, 9.85, -2.25, 7.75]}