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Math Help - can you help me with this please?

  1. #1
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    can you help me with this please?

    my problem is:

    "write an equation for the line parallel to the line -1/4y = -4 through the point (1, 8)"

    can you help me with this problem please? thanks
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  2. #2
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    When you are given a point (p,q) and a slope=m, you can construct the line with slope m through point (x,y) with the following:

    y-q=m(x-p)


    So you got m=2 and (p,q)=(-3,-5) so just plug it in

    Then you have to change it to y=mx+b, so once you plug your stuff in, multiply through on the right and add q to both sides

    y=mx-mp+q

    Can you make the correct substitutions?
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  3. #3
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    Quote Originally Posted by artvandalay11 View Post
    When you are given a point (p,q) and a slope=m, you can construct the line with slope m through point (x,y) with the following:

    y-q=m(x-p)


    So you got m=2 and (p,q)=(-3,-5) so just plug it in

    Then you have to change it to y=mx+b, so once you plug your stuff in, multiply through on the right and add q to both sides

    y=mx-mp+q

    Can you make the correct substitutions?
    yeah i got it right directly after i finished the problem

    for some reason i keep getting stuck at ones like these

    "write an equation for the line parallel to the line -1/4y = -4 through the point (1, 8)"

    i just cant seem to figure them out
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  4. #4
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    Well if we multiply each side by -4 you get y=16

    So the slope=0 cuz 16 is a constant,

    So you just need the equation y=8 for the line parallel that goes through (1,8)


    Let's make this general

    So they give us

    ay=b+cx

     <br />
y=\frac{b}{a}+\frac{c}{a}x<br />

    So we have found that the parallel slope= \frac{c}{a}

    Now we use the point slope form of the line to get (point is (q,p))

    y-p=\frac{c}{a}(x-q)

    Basically, always find the slope of the line given, then use the point thats given to write it as above
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  5. #5
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    ok, so i calculated it up, i got y=8 through the formula, is my answer correct?
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  6. #6
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    yep
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  7. #7
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    Quote Originally Posted by artvandalay11 View Post
    yep
    thank you very much sir
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  8. #8
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    there are various ways....the quick way is that in the equation you can see no mention of x and it is y equal to sumthing therefore it means it is a horizontal line with a gradient of 0
    therefore the line parallel would also be horizontal. therefore just take the y value of the point it is passing through and make its equal to y ie y=8

    the other way is changee the equation to the form y=mx+c where m is the gradient and c is the y intercept
    this is to find the gradient of the parallel line.
    y=(0)*(x) + 16
    therefor the value of m in the parallel line eq would be 0 as well
    y=(0)*(x) +c
    substitute the x and y values of the point it is passing through to find the value of c
    8=(0)*(1)+c
    c=8

    therefore you end up with y= (0)*(x) + 8
    which is the same as y=8
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