# How do you graph the line through point (2,6) and slope m=2/3?

Jun 22, 2018

See a solution process below:

#### Explanation:

First, we can plot the point given in the problem:

graph{((x-2)^2+(y-6)^2-0.1)=0 [-20, 20, -10, 10]}

Slope is $\text{rise"/"run}$ or the change in the $y$ values over the change in the $x$ values:

The slope in the problem is $\frac{2}{3}$/3#

Therefore,

The change in the $\text{rise}$ or $y$ value is $\textcolor{red}{+ 2}$

The change in the $\text{run}$ or $x$ value is $\textcolor{b l u e}{+ 3}$

We can find another point by adding these to the values from the point in the problem:

$\left(2 , 6\right) \to \left(2 \textcolor{b l u e}{+ 3} , 6 \textcolor{red}{+ 2}\right) \to \left(5 , 8\right)$

We can now plot this point:

graph{((x-2)^2+(y-6)^2-0.1)((x-5)^2+(y-8)^2-0.1)=0 [-20, 20, -10, 10]}

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

graph{(y-2/3x - 4.6)((x-2)^2+(y-6)^2-0.1)((x-5)^2+(y-8)^2-0.1)=0 [-20, 20, -10, 10]}