# Graphing an equation

• Apr 30th 2008, 06:15 PM
donnagirl
Graphing an equation
The graph of which of the following lines contains the point (5,0) and makes an angle of 45 degrees with the positive x-axis?

Now the solution is y=x-5 but why couldn't the solution be y=(3x-15)/2 or y=2x+10?? I guess I'm stuck on the the 45 degree part..
• Apr 30th 2008, 08:29 PM
mr fantastic
Quote:

Originally Posted by donnagirl
The graph of which of the following lines contains the point (5,0) and makes an angle of 45 degrees with the positive x-axis?

Now the solution is y=x-5 but why couldn't the solution be y=(3x-15)/2 or y=2x+10?? I guess I'm stuck on the the 45 degree part..

Formula: $m = \tan \theta$ where $\theta$ is the angle the line makes with the positive x-axis.
• Apr 30th 2008, 09:20 PM
donnagirl
Still am not sure...do you mean "m" is the slope?? And also why does it have to be tangent? Why not sin or cos?
• Apr 30th 2008, 10:42 PM
Gusbob
It is $tan \;\theta$ because it has been proven.

We know from coordinate geometry that the gradient of a straight line is the rise over run.

Consider the following triangle attached at the bottom of this post:
(Note: the diagonal line is "straight line" whose equation you are looking for)

If the gradient of the diagonal line is symbolised by the letter m

$m = \frac {rise}{run}$

$= \frac {opposite}{adjacent}$
From trigonometry, we know that $\frac {opposite}{adjacent}$ is also $tan \;\theta$

Therefore $m = tan\;\theta$

To find the equation of a straight line, you need a point $(x_1,y_1)$ and gradient m.