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

  1. #1
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    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?
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    Quote Originally Posted by Anemori View Post
    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.
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    Quote Originally Posted by masters View Post
    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?
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    Quote Originally Posted by Anemori View Post
    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,

    I can see how you might be confused about this problem.
    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)
    Last edited by masters; February 28th 2010 at 02:25 PM.
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    is is the same as y= mx + b and y-y_1=m(x-x_1)?
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    Quote Originally Posted by Anemori View Post
    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?
    Last edited by masters; February 28th 2010 at 02:17 PM.
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    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?
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    Quote Originally Posted by Anemori View Post
    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.
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  9. #9
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    Ok thanks!
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