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Math Help - Write an Equation? I dont know how to write an equation..

  1. #1
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    Write an Equation? I dont know how to write an equation..

    I'm in the 9th grade and i haven't been going to school.. I just started going again and my teacher gave me a ishload of homework to make up.. so i need help since i dont know how to do any of it..

    Write an equation of a line through the given point with the given slope.
    1. (3,-5);M = 2
    2. (1,2); M = -3
    3. (3,0); M = -1

    All help is appreciated.. Thanks in advance..
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  2. #2
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    Quote Originally Posted by JeeFong View Post

    Write an equation of a line through the given point with the given slope.
    1. (3,-5);M = 2
    Let me just give you a general formula.
    The equation of a line (non-vertical) is:
    y-y_0=m(x-x_0)
    Where,
    (x_0,y_0) a point on the line.
    Und m is the slope.

    Thus,
    y-(-5)=2(x-3)
    y+5=2x-6
    y=2x-5-5
    y=2x-10
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    Thanks .. but i still dont really get it.. could you break it down for me..
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    Quote Originally Posted by JeeFong View Post
    Thanks .. but i still dont really get it.. could you break it down for me..
    What do you not understand?
    The formula?
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    yea dont really understand it..
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    Quote Originally Posted by JeeFong View Post
    I'm in the 9th grade and i haven't been going to school.. I just started going again and my teacher gave me a ishload of homework to make up.. so i need help since i dont know how to do any of it..

    Write an equation of a line through the given point with the given slope.
    1. (3,-5);M = 2
    2. (1,2); M = -3
    3. (3,0); M = -1

    All help is appreciated.. Thanks in advance..
    M is the symbol we are using for the slope, which tells how "steep" the line is on the graph. (3, -5) is an "ordered pair" where the first number is the x value of the point and -5 is the y value of the point.

    Any line can be expressed in the form:
    y - y_0 = m(x - x_0)
    where (x_0, y_0) is a point on the line, m is the slope of the line, x is the independent variable, and y is the dependent variable.

    So for problem 1 we have that the slope m = 2, and (x_0, y_0) = (3, -5). Thus:
    y - (-5) = 2(x - 3)

    Now solve this for y:
    y + 5 = 2x - 6

    y = 2x - 11 (TPH had a slight typo in his example here.)

    The other two are done exactly the same way.

    -Dan
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  7. #7
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    Quote Originally Posted by JeeFong View Post
    yea dont really understand it..
    Hello,

    to set up the equation of a straight line you need 2 points: One fixed point P_0 with the coordinates x_0 and y_0 and a point P which represents every possible point of the straight line. It has the coordinates x and y.

    The slope of the line is calculated by the ratio of the legs of a right triangle m=\frac{red}{blue}. You can calculate the length of those legs:
    {red}=y-y_0
    {blue}=x-x_0
    That means:
    m=\frac{red}{blue}=\frac{y-y_0}{x-x_0}. Now multiply by the denominator:

    m \cdot (x-x_0)=y-y_0

    And that's exactly the formula TPH has used.

    EB
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  8. #8
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    Hello, JeeFong!

    Let me give you my baby-step approach . . . .


    Write an equation of a line through the given point with the given slope.

    1.\;\;(3,-5),\;m = 2

    If you are given a point on a line and the slope of the line,
    . . there is a formula we can use.
    It's called appropriately the "Point-Slope Formula".

    Given a point (x_1,y_1) and the slope m
    . . the equation of the line is: . y - y_1 \:=\:m(x - x_1)

    It's simply a matter of dropping in the facts we are given into the right slots.

    Look at the formula again: . y - y_1 \:=\:m(x - x_1)
    . . . . . . . . . . . . . . . . . . . . . . \uparrow\qquad\;\uparrow\qquad\;\;\;\uparrow
    . . . . . . . . . . . . . . . .
    y-coordinate . . slope . . . x-coordinate


    We are given: . (3,-5),\:m = 2
    . . . . . . . . . . . . \uparrow\quad\uparrow\qquad\quad\;\;\nwarrow
    . . . . . . .x-coordinate .y-coordinate . slope


    Drop these into the formula: . y - (-5) \:=\:2(x - 3)

    And simplify: . y + 5 \:=\:2x - 6\quad\Rightarrow\quad \boxed{y \:=\:2x - 11}

    Got it?

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