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maths help please

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    the line y= 1/2x +6 meets x axis at point c. find the equation of the line with gradient 2/3 that passed through point c.

    so x = -12

    y -6 = 2/3 (x+12)
    I x by 3 to get rid of fraction
    3y-18=2x+24

    now I have to put it in the form ax+by+c = 0

    the answer is 2x-3y + 24
    I got the 2x-3y but how do i get the 24?
    thanks
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    Point C is where the Line y=1/2x+6 meets the x axis

    You correctly rearranged to get x=-12, and since the x axis is where y=0, point C should be (-12,0)

    However, when you substituted this into y-y1=m(x-x1), you stated the y value as 6, when it should be 0 as that is the y-coordinate of point C.

    So you should get ---> y-0=2/3(x+12)

    y=2/3x + 24/3

    Multiply all values by 3 gives you--> 3y=2x+24

    Then rearrange to get--> 2x-3y+24=0
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    (Original post by DeclanCael)
    Point C is where the Line y=1/2x+6 meets the x axis

    You correctly rearranged to get x=-12, and since the x axis is where y=0, point C should be (-12,0)

    However, when you substituted this into y-y1=m(x-x1), you stated the y value as 6, when it should be 0 as that is the y-coordinate of point C.

    So you should get ---> y-0=2/3(x+12)

    y=2/3x + 24/3

    Multiply all values by 3 gives you--> 3y=2x+24

    Then rearrange to get--> 2x-3y+24=0
    Ahh thanks!! I was putting +6 as the y coord because its the y intercept loool but thats not even the same equation
    thanks
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    (Original post by samantham999)
    Ahh thanks!! I was putting +6 as the y coord because its the y intercept loool but thats not even the same equation
    thanks
    No problem
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    (Original post by DeclanCael)
    No problem
    for y=-2x +8 meets y axis at point b, find equation of line with gradient 2 passes through point B

    So I got y=2x-8 but it says y=2x+8
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    (Original post by samantham999)
    for y=-2x +8 meets y axis at point b, find equation of line with gradient 2 passes through point B

    So I got y=2x-8 but it says y=2x+8
    Point B is the y intercept, which you can get from the equation of the line as 8.

    The y intercept is when x=0, so Point B is at (0,8)

    Try putting this into the equation y-y1=m(x-x1) again, make sure you don't get your x's and y's mixed up.
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    (Original post by DeclanCael)
    Point B is the y intercept, which you can get from the equation of the line as 8.

    The y intercept is when x=0, so Point B is at (0,8)

    Try putting this into the equation y-y1=m(x-x1) again, make sure you don't get your x's and y's mixed up.
    I keep getting this y-8= 2(x-4)
    y-8=2x-8
    :/
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    (Original post by samantham999)
    I keep getting this y-8= 2(x-4)
    y-8=2x-8
    :/
    The left side of your answer is correct, so you are using the correct value for y. On the right side, you have the right gradient but not the correct x value.

    Remember that point B is (0,8) as we said before.
 
 
 
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