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In the form of y = m x + c.

  1. Mar 15, 2006 #1
    Hi All!

    I am trying to figure out how to solve problems like the one below. Any guidence would be appreciated.


    Work out the equation of the straight line through (3, -7) and (6, -16) in the form y = m x + c.

    m = (-7) - (-16) = 9 = -3
    __________ ___
    (3) - (6) -3

    m = -3

    y = mx + c

    -7 = (-3)(3) + c
    -7 = -9 + c
    c = -9 - (-7)
    c = -2

    y = (-3)x + -2

    That is what I have done so far, am I doing it correctly ?
     
  2. jcsd
  3. Mar 15, 2006 #2
    what's that about??

    also, you may find it easier to use the forumal y-y1=m(x-x1) because the formula just drops out of it
     
  4. Mar 15, 2006 #3

    Integral

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    Staff Emeritus
    Science Advisor
    Gold Member

    Just the night before last I was confronted by this exact problem at work. There are many different ways of finding the solution. There is also a very simple way of checking that you have the correct solution. Your final equation must be able to reproduce the given point when you plug them into your equation.

    What do you get when you do that? Do you get the correct answer? If not (you don't) then something is wrong.

    The way I approached the problem, was to write down the 2 equations which my given points provided. My points were

    (1,0) and (2, 25)

    so my equations were

    0 = m *1 + b
    and

    25 = m *2 +b

    I solved the first equation to get

    b = -m

    Using that in the second equation I got:

    25 = 2m -m or
    m = 25

    so my final line is

    y= 25x -25

    Does it work?

    let x = 1

    y= 25-25= 0 Yep that point is correct!

    Now the second point (2,25)

    y = 2*25 - 25 = 50 -25 = 25

    Sure enough, it works.

    Can you reproduce these steps with your numbers?
     
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