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    cn anyone plz help how to work out the following qs:
    1. A(-1,-2) B(7,2) and C(k,4), where k is constant, are the vertices of triangle ABC.
    Angle ABC is a right angle. the gradient of AB is 1/2x
    a) calculate the value of k

    I hv no clue
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    (Original post by shana_t)
    cn anyone plz help how to work out the following qs:
    1. A(-1,-2) B(7,2) and C(k,4), where k is constant, are the vertices of triangle ABC.
    Angle ABC is a right angle. the gradient of AB is 1/2x
    a) calculate the value of k

    I hv no clue
    The gradient of AB = 1/2.

    AB is perpendicular to BC.

    What is the gradient of BC?

    How are the gradients related?
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    My way may not be the best but it seems to work for this question:

    The gradient of AB= 1/2 and you said that it is a right angle triangle so you do the perpendicular so m2= -2

    i used the coordinates B(7,2) and C(K,4)

    made an equation 4-2/k-7=-2(gradient)
    rearrange:
    2=-2(k-7) 2=-2K+14 2K=12 K=6

    Hope that is explained well.
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    thnx for ur help, I got it
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    (Original post by shana_t)
    cn anyone plz help how to work out the following qs:
    1. A(-1,-2) B(7,2) and C(k,4), where k is constant, are the vertices of triangle ABC.
    Angle ABC is a right angle. the gradient of AB is 1/2x
    a) calculate the value of k

    I hv no clue
    As we have a right angle triangle, AB and BC perpendicular to one another. You have the gradient of AB as 1/2. You can work out the gradient of the perpendicular as 1/2 X m=-1 so m=-2 Now you know that the perpendicular to AB which is BC is a line with gradient -2. The equation of AC needs to be worked out. This is because you need the x value (k) at which y=4. You know that AC has gradient -2 and passes through point A, whose coordinates you already have as (-1, -2)
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    sorry again, I got stuck on this qs too:

    the straight line l1 passes through the points A and B with coordinates (0,-2) and (6,7)
    the equation of l1 is y=3/2x-2
    another straight line l2 with equation x+y=8 cuts the y-axis at point . The lines l1 and l2 intersect at point D (4,4)
    calculate the area of triangle ACD
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    (Original post by krisshP)
    As we have a right angle triangle, AB and AC are perpendicular to one another. You have the gradient of AB as 1/2. You can work out the gradient of the perpendicular as 1/2 X m=-1 so m=-2 Now you know that the perpendicular to AB which is AC is a line with gradient -2. The equation of AC needs to be worked out. This is because you need the x value (k) at which y=4. You know that AC has gradient -2 and passes through point A, whose coordinates you already have as (-1, -2)
    Angle ABC is the right angle meaning it is angle B not A so AB and BC are perpendicular
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    Angle C cuts the y axis meaning x=0 y+0=8 y=8

    A= (0,-2)
    C= (0,8)
    D=(4,4)

    AREA= BASE TIME HEIGHT/ 2

    Base= 10
    Height=4

    40/2 = 20
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    (Original post by simonb451)
    Angle C cuts the y axis meaning x=0 y+0=8 y=8

    A= (0,-2)
    C= (0,8)
    D=(4,4)

    AREA= BASE TIME HEIGHT/ 2

    Base= 10
    Height=4

    40/2 = 20
    thnxx a lot
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    (Original post by shana_t)
    thnxx a lot
    Thats fine- are you teaching yourself AS maths?
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    (Original post by simonb451)
    Thats fine- are you teaching yourself AS maths?
    yep, preparing for A level maths, gona start it from this september
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    (Original post by shana_t)
    yep, preparing for A level maths, gona start it from this september
    C1 and C2 are pretty good- Its S1 that most people seem to struggle with
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    (Original post by simonb451)
    C1 and C2 are pretty good- Its S1 that most people seem to struggle with
    ohh, my school does C1, C2 and D1. Wt about D1??
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    (Original post by shana_t)
    ohh, my school does C1, C2 and D1. Wt about D1??
    Oh we do D1 at A2, it sounds easy from what ive heard- the only tricky area of AS core was probably trig as it does take a while to get your head around sometimes
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    (Original post by simonb451)
    Oh we do D1 at A2, it sounds easy from what ive heard- the only tricky area of AS core was probably trig as it does take a while to get your head around sometimes
    oh okey
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    does anyone know how to work out this qs:
    the straight line l passes through A(1, 3√3) and B (2+√3,3+4√3)
    the gradient of l is √3
    give the equation of l in the form of y=mx+c, where constants m and c are surds given in their simplest form.
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    You can use y-y_1=m(x-x_1) or rearrange y=mx+c to get c=y-mx.

    Using the first option you will have y-3\sqrt{3}=\sqrt{3}(x-1) which you will need to rearrange.

    With the second option c=3\sqrt{3} -\sqrt{3}\times 1.
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    (Original post by BabyMaths)
    You can use y-y_1=m(x-x_1) or rearrange y=mx+c to get c=y-mx.

    Using the first option you will have y-3\sqrt{3}=\sqrt{3}(x-1) which you will need to rearrange.

    With the second option c=3\sqrt{3} -\sqrt{3}\times 1.
    the answer is y=√3x+2√3, I still don't get it
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    (Original post by shana_t)
    the answer is y=√3x+2√3, I still don't get it
    You have two points A(1, 3\sqrt3), B (2+\sqrt3,3+4\sqrt3) You've being told that the gradient of the line l which passes through these two points is m=\sqrt{3}

    Now the equation of a line is given by the formula y-y_{1}=m(x-x_{1}) This means that you can substitute any of the coordinates from either point to get the equation of l, so lets label A as the point A(x_{1},y_{1}) well from the question x_{1}=1 and y_{1}=3\sqrt3 for the coordinates of A correct?

    Substitute these into the formula along with m, expand and rearrange into the form y=mx+c
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    (Original post by shana_t)
    the answer is y=√3x+2√3, I still don't get it
    Here's a simpler example. A line with gradient 2 passes through the point (4,9). Find the equation of the line in the form y=mx+c

    You already know the equation is y=2x+c and you know that y=9 when x=4.

    You can use those values to find the value of c.

    9=2 \times 4 + c

    c=1

    y=2x+1
 
 
 
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