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Plotting Lines

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Plotting Lines

by MBA.Aspirant » Sat Jun 25, 2011 10:38 am
line x + 2y + 3 = 0 and line x - 2y + 3 = 0

line x+2y+3=0

y = -1/2x-3/2

x- intercept= 3

line x - 2y + 3 = 0

y= 1/2x +3/2

x-intercept = -3

Plotting line 2 is easy, but how to plot line 1? If it has a y-intercept -1.5 and x-intercept 3, then how the slope is -?

Thanks
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by Ashley@VeritasPrep » Sat Jun 25, 2011 10:52 am
MBA.Aspirant wrote:line x + 2y + 3 = 0 and line x - 2y + 3 = 0

line x+2y+3=0

y = -1/2x-3/2

x- intercept= 3

line x - 2y + 3 = 0

y= 1/2x +3/2

x-intercept = -3

Plotting line 2 is easy, but how to plot line 1? If it has a y-intercept -1.5 and x-intercept 3, then how the slope is -?

Thanks
Check your math on the x-intercept for line 1. Once you have the lines in slope-intercept form (y=mx+b), m is your slope and b is your y-intercept. So for your first line (y=(-1/2)x-(3/2)), your slope is -1/2 and your y-intercept is -3/2. Your x-intercept will occur when y = 0, so if you plug in 0 for y in that equation, you'll get 0 = (-1/2)x-(3/2) --> (1/2)x = -3/2 --> x = -3. That's your x-intercept there. (Note you don't actually NEED to find it, though, you can just plot your y-intercept and graph the line from there using the slope of -1/2... slope is rise over run, so you'll go down 1 (for the negative numerator) every time you move over to the right 2 (for the denominator).)


One alternative possibility: you can avoid entirely the step of putting the lines if slope-intercept form, if you like. Starting from the given line1: x+2y+3 = 0 and the given line2:x-2y+3=0, you can find the x-intercept of each by setting y to 0 and the y-intercept of each by setting x to 0. Plot the x- and y-intercepts for each line and just connect those dots!
Ashley Newman-Owens
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by Ashley@VeritasPrep » Sat Jun 25, 2011 11:01 am
Oh, and one other trick to have up your sleeve :)

Say you've got an equation in general form (ax + by = c), which is almost how your equations started out -- would just need to subtract the 3 in both cases to transport it to the other side of the equal sign.

Using the fact that the x-intercept occurs when y = 0, I can find the x-intercept as follows:
ax + b(0) = c ---> ax = c ---> x = c/a.

Using the fact that the y-intercept occurs when x = 0, I can find the y-intercept as follows:
a(0) + by = c ---> by = c ---> y = c/b.

Using the fact that the slope would be the coefficient of x if I were to translate this into general form, I can find it as follows:
ax + by = c ---> by = -ax + c ---> y = (-a/b)x + c/b.

So to summarize, given the equation of a line in general form ax + by = c,
your x-intercept is always c/a
your y-intercept is always c/b
your slope is always -a/b

I would consider it optional to memorize this, but it does save a couple steps sometimes!
Ashley Newman-Owens
GMAT Instructor
Veritas Prep

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