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

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1. 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)
2. Start off by drawing a labelled diagram of the triangle.

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.
3. (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
4. (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?????!
5. (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
6. (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.
7. (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!!

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