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New Maths 9-1 GCSE Sample Question HELP NEEDED!!? watch

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    My teacher gave me some practise questions for my end of year exam which will be like the new gcse and this question is very tackling to me. Could any with clear working solve the question and show me how they did it:

    A(-2,1), B(6,5), and C(4,k) are the vertices of a right angled triangle ABC.
    Angle ABC is the right angle.
    Find an equation of the line that passes through A and C.
    Give your answer in the form ay+bx=c where a, b and c are integers.
    (Total for Question = 5 marks)
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    Start off by drawing a labelled diagram of the triangle.

    Find the gradient of AB. Using this answer, what is the gradient of BC?

    How might you get the gradient of BC in terms of k? Once you've found the y coordinates of C it should all fall into place.
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    (Original post by NothingButWaleed)
    My teacher gave me some practise questions for my end of year exam which will be like the new gcse and this question is very tackling to me. Could any with clear working solve the question and show me how they did it:

    A(-2,1), B(6,5), and C(4,k) are the vertices of a right angled triangle ABC.
    Angle ABC is the right angle.
    Find an equation of the line that passes through A and C.
    Give your answer in the form ay+bx=c where a, b and c are integers.
    (Total for Question = 5 marks)
    Draw a diagram, it helps.
    From your diagram the only place where point C could be is vertically higher than A and C to create a right angled triangle

    Use the formula

    and use it on the points A and C, B and C since what you're doing is calculating the gradient you can say gradient= -gradient(because the other one is a negative gradient) then set equations equal to each other then solve for K then you're ok
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    (Original post by Big white)
    Draw a diagram, it helps.
    From your diagram the only place where point C could be is vertically higher than A and C to create a right angled triangle

    Use the formula

    and use it on the points A and C, B and C since what you're doing is calculating the gradient you can say gradient= -gradient(because the other one is a negative gradient) then set equations equal to each other then solve for K then you're ok
    Ok so I did:
    GRAD of AB = (5-1)/(6-(-2)) = 0.5
    GRAD OF CB = -1/0.5 = -2
    GRAD OF CB = (k-5)/(4-6) =
    -2/1 = (k-5)/-2 =
    k-5 = 4
    k = 9

    Sub C (4,9) into y-y1 = m(x-x1) =
    m = (9-1)/(4-(-2)) = 8/6
    y-9 = 8/6(x-4)
    6(y-9) * 8(x-4)
    AND NOW IM STUCK!!!
    I found the mark scheme and the answer is 6y-8x = 22, HOW?????!
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    (Original post by NothingButWaleed)
    Ok so I did:
    GRAD of AB = (5-1)/(6-(-2)) = 0.5
    GRAD OF CB = -1/0.5 = -2
    GRAD OF CB = (k-5)/(4-6) =
    -2/1 = (k-5)/-2 =
    k-5 = 4
    k = 9

    Sub C (4,9) into y-y1 = m(x-x1) =
    m = (9-1)/(4-(-2)) = 8/6
    y-9 = 8/6(x-4)
    6(y-9) * 8(x-4)
    AND NOW IM STUCK!!!
    I found the mark scheme and the answer is 6y-8x = 22, HOW?????!
    So this bit here is wrong
    you know now that C is (4,9) and that A is (-2,1)
    So use to find the gradient of the line AC

    then use again, this time you have the gradient so choose the point A or C to use and substitute the gradient and A or C into this and rearrange to find the answer
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    (Original post by NothingButWaleed)
    Ok so I did:
    GRAD of AB = (5-1)/(6-(-2)) = 0.5
    GRAD OF CB = -1/0.5 = -2
    GRAD OF CB = (k-5)/(4-6) =
    -2/1 = (k-5)/-2 =
    k-5 = 4
    k = 9

    Sub C (4,9) into y-y1 = m(x-x1) =
    m = (9-1)/(4-(-2)) = 8/6
    y-9 = 8/6(x-4)
    6(y-9) * 8(x-4)
    AND NOW IM STUCK!!!
    I found the mark scheme and the answer is 6y-8x = 22, HOW?????!
    You are pretty much there. Just expand out 6(y-9) = 8(x-4) and rearrange.
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    (Original post by SkyJP)
    You are pretty much there. Just expand out 6(y-9) = 8(x-4) and rearrange.
    Sometimes I wonder how stupid I am..
    It worked like a charm..
    6y-54=8x-32
    6y-8x-22!!
    Thanks for your help!!
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    Bland.in is a good site for Exam Styled Questions, however though, I would like more questions on this sort of maths to do with Graphs and finding the equation for a side of the shape etc. Is there any PDFs or sites that offer these questions?
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    (Original post by Big white)
    Draw a diagram, it helps.
    From your diagram the only place where point C could be is vertically higher than A and C to create a right angled triangle

    Use the formula

    and use it on the points A and C, B and C since what you're doing is calculating the gradient you can say gradient= -gradient(because the other one is a negative gradient) then set equations equal to each other then solve for K then you're ok
    this doesnt help ****. explain the formula and how tf you got the -1
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    (Original post by Clapped Pringles)
    this doesnt help ****. explain the formula and how tf you got the -1
    bro this was like 2 years ago, ive already done my GCSE exams like 7 months ago
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    (Original post by NothingButWaleed)
    bro this was like 2 years ago, ive already done my GCSE exams like 7 months ago
    How did you do?
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    (Original post by Y11_Maths)
    How did you do?
    Good, got an 8
    Was a surprise because at the start of year 11 I was getting 4s and 5s...stepped up my game big time near december!
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    (Original post by NothingButWaleed)
    Good, got an 8
    Was a surprise because at the start of year 11 I was getting 4s and 5s...stepped up my game big time near december!
    Nice well done
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