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Almost ashamed to ask this question.

  1. May 27, 2007 #1
    When we want to find the equation of a line we can use:
    [tex] Y-Y_1 = m(X-X_1)[/tex] or [tex] Y = mX + B[/tex]

    My question is aren't both equations essentially the same?

    For equation 1. If I know two points that the line passes through just by plugging and chugging won't that lead to what the Y intercept is?

    For example I have these two points that a line passes through: (2,12) & (6,0).
    by looking at the graph I know that the line has a Y intercept, even though it is not drawn. When I use equation one my constant (B) the Y intercept is 0 even though it should not be. What the heck am I doing wrong?:biggrin:
     
  2. jcsd
  3. May 27, 2007 #2

    cristo

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    Yes, they are the same equation. I'm not sure what you're doing wrong, but if we use your first equation, put one point in (say (6,0)), then y-0=m(x-6) => y=m(x-6). Then use the second point to find m; 12=m(2-6) => m=-3; thus y=-3(x-6)=-3x+18 is the equation for the line.
     
  4. May 27, 2007 #3
    Yes they are the same, the second one is a special case where [tex]Y_{1}[/tex] is the y intercept (B) and thus [tex]X_{1}[/tex] is 0 so you get Y=mX+[tex]Y_{1}[/tex]

    edit: woops, hi cristo =-).
     
  5. May 27, 2007 #4

    cristo

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    Haha; that gets me back for butting into your thread earlier! :wink:
     
  6. May 27, 2007 #5
    You have hit upon the point of my confusion exactly cristo! Jeez, so simple but I have not used this equation since precalc so i forgot how it worked!!! the point slope form of the line is used to find the slope of the line. It is the variable that we are solving for.
     
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