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pure core 1 maths chap 3 - coordinate geometry of straight lines question Watch

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    Find, in the form ax+by+c=0, the equation of the line with gradient 2 and y-intercept -3
    I get that you're supposed to rearrange it into the form y=mx+c, but that only gives the values of a/b and c/b, not the letters by their selves.
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    (Original post by daviem)
    Find, in the form ax+by+c=0, the equation of the line with gradient 2 and y-intercept -3
    I get that you're supposed to rearrange it into the form y=mx+c, but that only gives the values of a/b and c/b, not the letters by their selves.
    For the simplest integer coefficients in that form, a = -2, b = 1, c = 3. Any multiples of this would also work, but it's best to simplify fully.
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    (Original post by daviem)
    Find, in the form ax+by+c=0, the equation of the line with gradient 2 and y-intercept -3
    I get that you're supposed to rearrange it into the form y=mx+c, but that only gives the values of a/b and c/b, not the letters by their selves.
    All the question is asking you to do is sub in your values of m and c, then rearrange y = 2x-3 into the form 0 = 2x-y-3

    You might be over complicating it
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    (Original post by morgan8002)
    For the simplest integer coefficients in that form, a = -2, b = 1, c = 3. Any multiples of this would also work, but it's best to simplify fully.
    Could you show me how you did it
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    (Original post by daviem)
    Could you show me how you did it
    You can rearrange the equation that you had:  y = -\frac{a}{b}x -\frac{c}{b}
    Then \frac{a}{b} = -2, \frac{c}{b} = 3
    Settting b = 1 for simplification, a = -2, c = 3

    Or you can go from y = 2x - 3 and rearrange to -2x + y + 3 = 0
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    (Original post by Actaeon)
    All the question is asking you to do is sub in your values of m and c, then rearrange y = 2x-3 into the form 0 = 2x-y-3

    You might be over complicating it
    But why does a=2 when we work -a/b out as being 2?
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    (Original post by morgan8002)
    You can rearrange the equation that you had:  y = -\frac{a}{b}x -\frac{c}{b}
    Then \frac{a}{b} = -2, \frac{c}{b} = 3
    Settting b = 1 for simplification, a = -2, c = 3

    Or you can go from y = 2x - 3 and rearrange to -2x + y + 3 = 0
    So you're supposed to take b to equal 1 even though the question doesn't say that?
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    (Original post by daviem)
    But why does a=2 when we work -a/b out as being 2?
    Using the straight line equation y=mx+c

    m is the gradient, so y=2x+c
    c is the y intercept, so y=2x-3
    Rearrange to give 0=2x-y-3
    Using the y=mx+c equation is a more intuitive way of doing things I find
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    (Original post by daviem)
    So you're supposed to take b to equal 1 even though the question doesn't say that?
    Yeah. In this situation, you can pick any(non zero) value you want for one of the variables, and the other two will change their values to match. I picked 1 because it's simple. If you don't like that, just rearrange the y = mx + c into the form you want.
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    (Original post by morgan8002)
    Yeah. In this situation, you can pick any(non zero) value you want for one of the variables, and the other two will change their values to match. I picked 1 because it's simple. If you don't like that, just rearrange the y = mx + c into the form you want.
    Oh ok, I just didn't realise that was allowed, thanks for your help
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    (Original post by morgan8002)
    Yeah. In this situation, you can pick any(non zero) value you want for one of the variables, and the other two will change their values to match. I picked 1 because it's simple. If you don't like that, just rearrange the y = mx + c into the form you want.
    Ok, I understand that, thanks for the help
 
 
 
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