# How do you graph the lines using slope-intercept form y = -2/3x + 1?

Nov 27, 2017

See a solution process below:

#### Explanation:

This equation is in slope intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

$y = \textcolor{red}{- \frac{2}{3}} x + \textcolor{b l u e}{1}$

Therefore, the y-intercept is: $\textcolor{b l u e}{b = 1}$ or $\left(0 , \textcolor{b l u e}{1}\right)$

We can plot this point on the grid as:

graph{(x^2 + (y-1)^2 - 0.025) = 0 [-10, 10, -5, 5]}}

The slope is: $\textcolor{red}{m = - \frac{2}{3}}$

Slope is also: $\text{rise"/"run}$

In this case, the $\text{rise}$ is $- 2$ so we need to go down 2 positions on the $y$ value. And, the run is $3$ so we need to go right 3 positions on the $x$ value.

This second point is: $\left(0 + 3 , 1 - 2\right) \implies \left(3 , - 1\right)$

We can now plot this point:

graph{(x^2 + (y - 1)^2 - 0.025)((x - 3)^2 + (y + 1)^2 - 0.025) = 0 [-10, 10, -5, 5]}}

Now, we can draw a line through the two points giving:

graph{(y + (2/3)x - 1)(x^2 + (y - 1)^2 - 0.025)((x - 3)^2 + (y + 1)^2 - 0.025) = 0 [-10, 10, -5, 5]}}