# How do you graph x+y=5 using intercepts?

Jan 3, 2017

1. Get the $x$-intercept by setting $y = 0$ and solving for $x$.
2. Get the $y$-intercept by setting $x = 0$ and solving for $y$.

Graph these two points, and connect them.

#### Explanation:

An intercept is simply a point where a line (or any function) crosses an axis. At such a point, the non-axis coordinate is 0. (e.g. all points on the $x$-axis have no displacement in the $y$ direction, so the $y$-coordinate is $0$.)

We can use this to help us get two points that are on the line, graph those two points, and finally connect them by drawing a line through them.

The $x$-intercept occurs when $y = 0$. To find this intercept, we just need to plug in 0 for $y$ in our line equation:

$\textcolor{w h i t e}{\implies} x + y = 5 \text{ @ } y = 0$
$\implies x + 0 = 5$
$\implies x \textcolor{w h i t e}{+} \textcolor{w h i t e}{0} = 5$

So, when $y = 0$, we have $x = 5$, meaning $\left(5 , 0\right)$ is a point on our line, and it is the $x$-intercept.

Similarly, to find the $y$-intercept, we let $x = 0$ and solve for $y$:

$\textcolor{w h i t e}{\implies} x + y = 5 \text{ @ } x = 0$
$\implies 0 + y = 5$
$\implies \textcolor{w h i t e}{x} \textcolor{w h i t e}{+} y = 5$

So when $x = 0$, $y = 5$, and thus $\left(0 , 5\right)$ is our $y$-intercept.

Finally, we graph these two points, and connect them with a line:

graph{(x+y-5)((x-5)^2+y^2-0.06)(x^2+(y-5)^2-0.06)=0 [-8.325, 11.68, -2.3, 7.7]}

## Bonus:

For a linear equation of the form $x + y = k$ for some number $k$, the equation is saying "two numbers sum to $k$". From this, it is easy to see that when either number is 0, the other one must be $k$, so that the sum will be $k$. This means the intercepts will always be $\left(0 , k\right)$ and $\left(k , 0\right)$.