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Math Help - The Straight Line Question

  1. #1
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    The Straight Line Question

    Ques: Find the equation of the line that passes through the y-intercept of the line 3x+5y+2=0 and is parallel to the line 2x-4y-1=0.


    From 3x+5y+2=0 , i found out that y-intercept= -2/5 x-intercept= -2/3

    From 2x-4y-1=0 the, i found out that m= 1/2

    Since m1=m2, i sub into the formula: y+2/5=1/2(x+2/3)
    answer i got is y=1/2x-1/15 but is wrong.


    can anybody help? Thanks.
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  2. #2
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    Re: The Straight Line Question

    Quote Originally Posted by freshbox View Post
    Ques: Find the equation of the line that passes through the y-intercept of the line 3x+5y+2=0 and is parallel to the line 2x-4y-1=0.
    From 3x+5y+2=0 , i found out that y-intercept= -2/5 x-intercept= -2/3 From 2x-4y-1=0 the, i found out that m= 1/2
    Why are you doing anything with the x-intercept?

    The new line has the same y-intercept, \left(0,\tfrac{-2}{5}\right) as the line 3x+5y+2=0
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    Re: The Straight Line Question

    I am trying to find the x-intercept and y-intercept to use it in this formula: y-y1=m(x-x1)
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  4. #4
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    Re: The Straight Line Question

    Quote Originally Posted by freshbox View Post
    I am trying to find the x-intercept and y-intercept to use it in this formula: y-y1=m(x-x1)
    You can still use that formula: x_1=\tfrac{-2}{5}~\&~y_1=0.
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  5. #5
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    Re: The Straight Line Question

    How come your x1= -2/5 y1= 0 ?

    I thought i found the intercept as y-intercept= -2/5 x-intercept= -2/3

    Shouldn't y1=-2/5 and x1=-2/3?


    I know i am wrong, please explain it to me if you don't mind, thanks.
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  6. #6
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    Re: The Straight Line Question

    Quote Originally Posted by freshbox View Post
    How come your x1= -2/5 y1= 0 ?
    I thought i found the intercept as y-intercept= -2/5 x-intercept= -2/3
    Shouldn't y1=-2/5 and x1=-2/3? I know i am wrong, please explain it to me if you don't mind, thanks.
    If you write the equation of a line through (p,q) with slope m you get y-q=m(x-p) (point-slope form).

    So line through \left(0,\tfrac{-2}{5}\right) with slope \tfrac{1}{2} is y+\tfrac{2}{5}\right=\tfrac{1}{2}\left(x-0).
    Last edited by Plato; December 10th 2011 at 06:36 AM.
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  7. #7
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    Re: The Straight Line Question

    i don't understand
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    Re: The Straight Line Question

    (p,q) as (x,y)
    (p,q) y-q=m(x-p)

    this part i get it

    (x,y)
    (0,-2/5) y-0=1/2(x+2/5)

    this part how come your x and y swap places? i thought the formula is : y1-y=m(x1-x)
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  9. #9
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    Re: The Straight Line Question

    You are correct. What we get for copy and paste.
    I edited it and it is now correct.
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  10. #10
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    Re: The Straight Line Question

    if the question says passes through the y-intercept means x-intercept = 0? Is this always the case?

    So the x-intercept i calculated as -2/3 is wrong?
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  11. #11
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    Re: The Straight Line Question

    Rewrite in slope-intercept form is what I would do first, personally, for this type of problem.

    y=(-3/5)x-2/5
    y=(1/2)x-1/4

    Since it'll be parallel to equation 2, y=mx+b m=1/2.


    Now, with x=0 in the first equation, the y-intercept, y=(-3/5)(0)-2/5=-2/5. So y=(1/2)x+b must equal -2/5 when x=0. -2/5=b.

    y=(1/2)x-2/5, or 5x-10y-4=0.
    Last edited by Feryll; December 10th 2011 at 07:14 PM.
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  12. #12
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    Re: The Straight Line Question

    if the question says passes through the y-intercept means x-intercept = 0? Is this always the case?
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  13. #13
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    Re: The Straight Line Question

    You aren't supposed to be paying attention to x-intercepts in this question. The y-intercept is the point on the graph where the line intersects the vertical y-axis, which is at x=0, no? Therefore, to find where the graph passes through the y-axis, you must set x=0 and evaluate the equation, which will return the y-intercept.

    (Just for reference, the x-intercept is solved by substituting 0 in place of y and solving for x, but that isn't necessary for this problem)
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