# A triangle has corners with coordinates (1,2), (-2,3) and (0,-1). Please help me determine the equations of the lines.

A triangle has corners with coordinates (1,2), (-2,3) and (0,-1). Please help me determine the equations of the lines that form the sides of the triangle.

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So which edge did you start on and where are you stuck? – rschwieb Mar 18 '13 at 16:58
I drew it out on a graph but am unsure what to do. – Dillan Mar 18 '13 at 16:59

The formula of a line is $y = mx+b$. You have two $(x,y)$ pairs for each line. I'll give you the first one:

$y = mx+b$

$2 = m+b$

$3 = -2m+b$

$m+b-2 = -2m+b-3$

Solve for $m$:

$m = -2m-1$

$3m = -1$

$m = -\frac{1}{3}$

Plug that back in for $b$:

$2 = m+b$

$2 = -1/3+b$

$b = 2\frac{1}{3}$

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thanks a lot that helps – Dillan Mar 18 '13 at 17:07

Do you know the point-slope equation of a line? If a line has slope $m$ and $(a, b)$ is a point on that line then the equation for the line is $$y - b = m(x - a)$$

You can find the slope of the line through, for example, $(1, 2)$ and $(-2, 3)$ by calculating the rise/run: $$m = \frac{3 - 2}{-2 - 1} = -\frac{1}{3}$$ so my slope is $-\frac{1}{3}$ and I choose point $(1, 2)$ and I get the following equation for the first line: $$y - 2 = -\frac{1}{3}(x - 1)$$ I'll leave the others for you.

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To find the equation of a line, we need a point and a slope. Given two points $(x_1,y_1)$ and $(x_2,y_2)$, we can determine the slope between them by using $$m=\frac{y_2-y_1}{x_1-x_2}$$ Next, use the point slope formula $$y-y_1=m(x-x_1)$$

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But this is three points in a triangle – Dillan Mar 18 '13 at 17:01
You're absolutely right. Notice, however, that any two vertices of the triangle define a side. Since there are three sides, we need to find the equations of three lines. Above is a suggested way to find one of these equations starting with a pair of vertices. To solve the problem, one must do this for all three sides, finding three equations. – Jared Mar 18 '13 at 17:03
okay thanks so for a parallelogram would it be work out four equations? – Dillan Mar 18 '13 at 17:04
You got it! The method for each equation would be exactly the same though; you're just starting with a different pair of vertices. Notice that for a parallelogram, opposite sides have equal slope, so that should save you some work. – Jared Mar 18 '13 at 17:06
okay thanks very much – Dillan Mar 18 '13 at 17:08

What have you tried? Begin by drawing a picture (helpful) and computing some slopes.

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i have thanks for your help – Dillan Mar 18 '13 at 17:07
I am trying to avoid giving away the question and am attempting to get the poser to think further about this question. Wny the downvote? – ncmathsadist Mar 18 '13 at 18:22
I was not the one who downvoted you, but I suspect the reason is that this seems more appropriate as a comment. – Jim Mar 18 '13 at 20:31
It is a partial answer. I am trying to give a hint that will help the OP over the lip without giving away the question. – ncmathsadist Mar 22 '13 at 1:11