# Math Help - Finding Equation for 3 lines passing through the vertices.

1. ## Finding Equation for 3 lines passing through the vertices.

Hello! I'm confused... I know how to find the equation for two points/lines but I don't know what formula should I use for finding equations for 3 lines. should it be the same and i have to find each line and combine them?

2. Originally Posted by Anemori
Hello! I'm confused... I know how to find the equation for two points/lines but I don't know what formula should I use for finding equations for 3 lines. should it be the same and i have to find each line and combine them?
Hi Anemori,

Just state the problem you have and we'll help you solve it.

3. Originally Posted by masters
Hi Anemori,

Just state the problem you have and we'll help you solve it.
Ok! Here is the problem:

I draw a Triangle with a vertices (0,8),(12,-8) and (-12,-8) on a graph.

Q: Find the equations for the three lines passing through the vertices.

i know how to get a equation of slope-intercept by using y=mx + b

but getting eqauation for those 3 lines... how?

4. Originally Posted by Anemori
Ok! Here is the problem:

I draw a Triangle with a vertices (0,8),(12,-8) and (-12,-8) on a graph.

Q: Find the equations for the three lines passing through the vertices.

i know how to get a equation of slope-intercept by using y=mx + b

but getting eqauation for those 3 lines... how?
Hello Anemori,

It's not really worded in the best way.

Here are your vertices: A(0, 8), B(12, -8), C(-12, -8).

You need to define the equations of the lines AB, BC, and AC.

Can you do that using the point-slope form of the linear equation: $y-y_1=m(x-x_1)$

5. is is the same as y= mx + b and $y-y_1=m(x-x_1)$?

6. Originally Posted by Anemori
is is the same as y= mx + b and $y-y_1=m(x-x_1)$?
$y-y_1=m(x-x_1)$ is called the 'point-slope form' of a linear equation.

If you don't know this one, and you haven't been told to use it, then we can use the $y=mx+b$

You'll just have to find the slope and y-intercept of each line.

So, which way do you want to go?

7. These are the equation i got for

Line AB $y= \frac {4}{3}x + 8$

Line BC $y= 0x + (-8)$

Line AC $y= - \frac {4}{3}x + 8$

What should i do next?

are those am i looking for?

8. Originally Posted by Anemori
These are the equation i got for

Line AB $y= \frac {4}{3}x + 8$

Line BC $y= 0x + (-8)$

Line AC $y= - \frac {4}{3}x + 8$

What should i do next?

are those am i looking for?
According to the vertices being A(0, 8), B(12, -8), C(-12, -8), I believe you need to switch equations AB and AC.

Line AB has the negative slope and line AC has the positive slope.

After that, you're done.

Line BC is y = -8. That is correct.

9. Ok thanks!