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

##### 1 Answer
Oct 30, 2015

Refer to the explanation.

#### Explanation:

Use the point-slope form for a linear equation. The general equation is $y - {y}_{1} = m \left(x - {x}_{1}\right)$, where ${x}_{1}$ and ${y}_{1}$ are the known point, and,$m$ is the slope.

$\text{Known point} = \left(0 , 4\right)$
${x}_{1} = 0$
${y}_{1} = 4$
$m = - \frac{2}{3}$

Substitute the known values and solve for $y$.

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 4 = - \frac{2}{3} \left(x - 0\right) =$

Add $4$ to both sides.

$y = - \frac{2}{3} \left(x - 0\right) + 4$

Distribute $- \frac{2}{3}$.

$y = - \frac{2}{3} \cdot x - \left(- \frac{2}{3} \cdot 0\right) + 4$

Simplify.

$y = - \frac{2}{3} x + 0 + 4 =$

$y = - \frac{2}{3} x + 4$

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

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