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Math Help - Problems on Slope

  1. #1
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    Problems on Slope

    Here's the problem:

    Problem: Recall that you can use two points on a line to determine an equation for the line. Suppose two points in a coordinate plane have coordinates (x1, y1) and (x2, y2).
    A. Write a formula that gives the slope b of the line containing the two points.
    Alright, I know this is b = \frac{y1-y2}{x1-x2}. No problem there.
    B. Solve the equation for in Part a for y1 in terms of the other variables.
    I don't quite remember how to do this. In the end, I got Y1 = b(x1-x2)+y2 Please correct me on this.
    C. In your equation from Part b, replace the point with coordinates (x1, y1), with a general point (x, y) on the line. Explain how this equation is now that of a line with slope b through the point (x2, y2).
    This is the part I thought was a brain teaser/tongue twister. What do you do here?
    D. Use your equation from Part c to write the equation of the line that contains the points (8, 12) and has slope -2. What is the y-intercept of this line?
    I don't have the equation for C. yet, so I'll wait for help on that first.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by PandaPanda View Post
    Here's the problem:

    Alright, I know this is b = \frac{y1-y2}{x1-x2}. No problem there.
    yup

    I don't quite remember how to do this. In the end, I got Y1 = b(x1-x2)+y2 Please correct me on this.
    that's fine

    This is the part I thought was a brain teaser/tongue twister. What do you do here?
    do as they say. change (x1, y1) to (x,y). simplify to get the equation in the slope-intercept form (y = mx + b, here, m is the slope and b is the y-intercept). you will be able to give them the info they require from this form

    I don't have the equation for C. yet, so I'll wait for help on that first.
    well, i gave you the hint on how to do C, so now you can try again
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  3. #3
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    Alright, thanks for clarifying.
    However, in Y = b(x-x)+y, the x's cancel out. How do I change that?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by PandaPanda View Post
    Alright, thanks for clarifying.
    However, in Y = b(x-x)+y, the x's cancel out. How do I change that?
    remember, one of the x's is x1 and the other is x2. they are not the same, you don't change both of them, just x1
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  5. #5
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    C) When you do the sub. you get y=b(x-x_{2})+y_{2}

    This is obviously still a linear equation, so nothing weird here. You can confirm quickly that this still has a slope of b as when you solve for b you get the slope between the two points (x,y) and (x2,y2). If you recall the Point-Slope form of a linear equation, the form of this last equation fits it exactly. Look Linear equation - Wikipedia, the free encyclopedia there to brush up.
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  6. #6
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    Ah, okay. Just one last question, because I'm slow :P

    After expanding, the result is:
    Y = bx-bx2+y2
    How do you deal with -bx2? Pretty sure it's not supposed to be there.

    Thanks very much. ~Panda
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by PandaPanda View Post
    Ah, okay. Just one last question, because I'm slow :P

    After expanding, the result is:
    Y = bx-bx2+y2
    How do you deal with -bx2? Pretty sure it's not supposed to be there.

    Thanks very much. ~Panda
    write as y = bx + (y_2 - bx_2)

    that is in the form y = mx + b, your (y_2 - bx_2) is just a constant
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  8. #8
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    Ok so use the equation y=b(x-x_{2})+y_{2}. You have the point (8,12) and a slope of -2. Plug in (8,12) for (x2,y2) and -2 for b.

    y=-2(x-8)+12 Simplifying this and moving towards a form of y=mx+b you get y=-2x+16+12=-2x+28. Now do you know how to get the y-intercept from here?
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