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    I am currently doing C1 revision for chapter 5 (Coordinate geometry in the x,y plane) and had a slight issue with the format of answers. when giving an answer in the form ax+by+c, muxt a always be a positive integer?
    With giving answers in the form y=mx+c, does then y have to be a positive, integer?
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    for ax+by+c=0 a can be positive or negative. the question will probably specify but if you get a to be negative and you multiply both sides by -1 a becomes positive. You'll still get the marks.

    in y=mx+c y isn't a value, only m and c can be integers (unless of course you put points in)
    (it can be positive or negative too)
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    (Original post by BackLumbarJack)
    I am currently doing C1 revision for chapter 5 (Coordinate geometry in the x,y plane) and had a slight issue with the format of answers. when giving an answer in the form ax+by+c, muxt a always be a positive integer?
    With giving answers in the form y=mx+c, does then y have to be a positive, integer?
    Basically, no it doesnt but when you're writing ax+by+c it would be most helpful if a is positive but co-ordinates can be positive or negative it doesn't really matter. show an example?
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    (Original post by CorpusLuteum)
    Basically, no it doesnt but when you're writing ax+by+c it would be most helpful if a is positive but co-ordinates can be positive or negative it doesn't really matter. show an example?
    http://pmt.physicsandmathstutor.com/...ht%20lines.pdf

    so for question 2b here, I got 15x-2y-50=0, as this is how the textbook advises we leave our answers, but the mark scheme gives something completely different; would I get the marks?
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    (Original post by BackLumbarJack)
    http://pmt.physicsandmathstutor.com/...ht%20lines.pdf

    so for question 2b here, I got 15x-2y-50=0, as this is how the textbook advises we leave our answers, but the mark scheme gives something completely different; would I get the marks?
    What you got, 15x-2y-50=0, is equivalent to the answer, 2y-15x+50 = 0.

    One has been multiplied by '-1' to get the other, but they both describe the same graph.
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    (Original post by K-Man_PhysCheM)
    What you got, 15x-2y-50=0, is equivalent to the answer, 2y-15x+50 = 0.

    One has been multiplied by '-1' to get the other, but they both describe the same graph.
    cheers man, so if i gave answers in this format I would definitely gain the marks?
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    (Original post by BackLumbarJack)
    cheers man, so if i gave answers in this format I would definitely gain the marks?
    The mark scheme says "o.e." next to the answer, which means "or equivalent", and as yours is equivalent to the answer, yes, you get all the marks.

    "o.e." is usually implied anyway, even if not explicitly stated on mark schemes (unless the question is very particular about exactly what form it wants it in, ie if a question wants the answer in the form k \sqrt{3} and you write 3 \sqrt{12} instead of 6 \sqrt{3}, then you won't gain the final mark).
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    (Original post by K-Man_PhysCheM)
    The mark scheme says "o.e." next to the answer, which means "or equivalent", and as yours is equivalent to the answer, yes, you get all the marks.

    "o.e." is usually implied anyway, even if not explicitly stated on mark schemes (unless the question is very particular about exactly what form it wants it in, ie if a question wants the answer in the form k \sqrt{3} and you write 3 \sqrt{12} instead of 6 \sqrt{3}, then you won't gain the final mark).
    how about for question 3b, if I put 3y=5x-12?
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    (Original post by BackLumbarJack)
    how about for question 3b, if I put 3y=5x-12?
    There you are told to leave it in the form y = mx +c, where m and c are constants. There should be no constant in front of the y. You would not get the final accuracy mark. The mark scheme states it must be left in that form.

    It may seem harsh, but this is convention because in this form (with no coefficient in front of the y) you can immediately tell what the gradient and y-intercept are from looking at the equation.
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    (Original post by K-Man_PhysCheM)
    There you are told to leave it in the form y = mx +c, where m and c are constants. There should be no constant in front of the y. You would not get the final accuracy mark. The mark scheme states it must be left in that form.

    It may seem harsh, but this is convention because in this form (with no coefficient in front of the y) you can immediately tell what the gradient and y-intercept are from looking at the equation.
    thanks
 
 
 
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