# Finding any point on a line if you know the slope and $y$-intercept.

I am wondering if there is a way to determine where a point is if I only know the slope and $y$-intercept.

For example, say I am told that the line has a slope of $3$ and a $y$-intercept of $-3$. How would I know how exactly to plot or draw the line? (Hope this makes sense).

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If we denote any point on the line as $(x,y)$, then all the points on the line satisfy the equation $$y = mx+c$$ where $m$ is the slope and $c$ is the $y$ intercept. This is called the equation of the line.

In your case, the slope is $m=3$ and the $y$ intercept is $c=-3$. Hence the equation is $$y = 3x-3$$

Below is the plot. The plot was generated using grapher on mac.

A simple way to plot is to identify the $x$ intercept and the $y$ intercept. You are given that the $y$ intercept is $-3$. To find the $x$ intercept, set $y = 0$. This gives you that $3x-3 = 0$ i.e. $x = 1$. Hence, you know that the line passes through $(1,0)$ and $(0,-3)$. Join these two points and extend them on both sides.

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This is exactly what I was wondering, but I'm a little confused as to how 3x - 3 makes x = 1. Sorry for the ignorant question but could you clarify? – daveMac May 24 '12 at 23:18
@daveMac To get the $X$ intercept, what you need to do is to set $y = 0$ in the equation of the line. The equation of the line is $y = 3x-3$. Setting $y=0$, gives us $0 = 3x -3$. Adding $3$ to both sides, we get that $3 = 3x$. Dividing by $3$ on both sides, gives us $x = \frac3{3} = 1$. This is the $x$ intercept. Hope it is clear now. – user17762 May 24 '12 at 23:22
Thanks Marvis...I'm a bit fuzzy on my algebra. That cleared it right up. – daveMac May 24 '12 at 23:28

Marivs has given a detailed explanation of how to this algebraically. Here is a slightly different way of looking at it from the definitions of slope and $y$-intercept:

You are given the $y$-intercept is -3. From the definition of $y$-intercept a point on the graph is $(0,-3)$.

The defintion of slope is $$\text{slope} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$$ where $(x_1,y_1)$ and $(x_2,y_2)$ are any two points on the line.

So, if you let $(x_1,y_1) = (0,-3)$, and using the fact that the slope is 3, then any other point on the graph can be obtained by $$3 = \frac{y - (- 3)}{x - 0}$$ so $$y = 3x - 3$$

Now, just pick any value of $x$ (other than $0$, since we already have this point, which is the $y$-intercept), and you obtain a second point on the line. Say we let $x=1$, then $y=0$. So connecthing this point $(1,0)$ to the given point $(0,-3)$ you get the graph of this line.

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This explanation really helped me as well...I was torn on which answer to accept. Thanks for the effort. – daveMac May 24 '12 at 23:42

If you want to draw it easily, one thing that you can do is just find anouther point on the line. It dosen't matter which, just as long as it is definitly not the y-intercept given. Then, take a ruler and connect the lines and extend indefinitly.

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Since I interpret the question to be asking how to find another point on the line, this doesn't seem to answer the question. – Ronald May 24 '12 at 23:05