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Math Help - Graph; have gradient, need y intercept

  1. #1
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    Graph; have gradient, need y intercept

    I need to find the equation for a straight line that goes through point (2,4) and has a gradient of 3

    I know the equation is y=mx+c but I don't have c. If I treat the point I was given as c (c=4) and add 2 to x will it be right?

    y=3(x+2)+4

    Or are the brackets messing it up? It doesn't look like it would be right. Could it be;

    y=3x+6

    That also doesn't seem right. I don't know.

    Thanks in advance!
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  2. #2
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    Re: Graph; have gradient, need y intercept

    No, c is the y-intercept. It should be

    y = mx + c where x = 2, y = 4, m = gradient = 3

    4 = 2(3) + c \Rightarrow 4 = 6+c

    c = -2
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  3. #3
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    Re: Graph; have gradient, need y intercept

    Another way to find the eqn. of a line when you don't have the y-intercept is to use the eqn. y-b=m(x-a) where (a,b) is a point on the line and m is the gradient.
    e.g.
    m=3, pt (2,4) so,
    a=2, b=4
    y-b=m(x-a)
    y-4=3(x-2)
    y-4=3x-6
    y=3x-2

    I prefer to use y=mx+c when I have the y-intercept, but y-b=m(x-a) when I don't.
    Also the latter eqn. is used more in the Higher course in Scotland.
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  4. #4
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    Re: Graph; have gradient, need y intercept

    Quote Originally Posted by Badgers View Post
    I need to find the equation for a straight line that goes through point (2,4) and has a gradient of 3

    I know the equation is y=mx+c but I don't have c. If I treat the point I was given as c (c=4) and add 2 to x will it be right?

    y=3(x+2)+4
    You can check that this is incorrect by setting x= 2: y= 3(2+ 2)+ 4= 12+ 4= 16, not 4. The formula you were trying to remember is y= m(x- x_0)+ y_0. You might aso recall that the slope of a line between two points (x_1, y_1) and (x_2, y_2) is [tex]\frac{y_1- y_2}{x_1- x_2}[/itex]. If you know that the slope is 3 and that one point on the line is (2, 4), then that means that for a "generic" point, (x, y), 3= \frac{y- 4}{x- 2} so that y- 4= 3(x- 2) so that y= 3(x- 2)+ 4 as I said above.

    Another way of doing this problem is to note that the slope is "rise over run". With x going from 2 to 4, the "run" is -4 so the "rise" is 3(-4)= -12. y goes from 2 down to 2- 12= -10.
    Or are the brackets messing it up? It doesn't look like it would be right. Could it be;

    y=3x+6
    Again, if x= 2, y= 3(2)+ 6= 12, not 4 so that can't be right.

    That also doesn't seem right. I don't know.

    Thanks in advance!
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