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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NothingButWaleed
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- 11-06-2016 17:28
Last edited by NothingButWaleed; 11-06-2016 at 17:44. -
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- 11-06-2016 18:06
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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- 11-06-2016 18:14
(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)
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 okLast edited by Big white; 11-06-2016 at 18:16. -
NothingButWaleed
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- 12-06-2016 12:27
(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
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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- 12-06-2016 12:38
(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 know now that C is (4,9) and that A is (-2,1)
So useto find the gradient of the line AC
then useagain, 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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- 12-06-2016 14:02
(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?????! -
NothingButWaleed
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- 12-06-2016 15:03
(Original post by SkyJP)
You are pretty much there. Just expand out 6(y-9) = 8(x-4) and rearrange.
It worked like a charm..
6y-54=8x-32
6y-8x-22!!
Thanks for your help!! -
JosephCiderBwoy
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- 09-01-2017 17:45
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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Clapped Pringles
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- 02-03-2018 19:44
(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 -
NothingButWaleed
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- 04-03-2018 13:17
(Original post by Clapped Pringles)
this doesnt help ****. explain the formula and how tf you got the -1 -
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- 04-03-2018 16:52
(Original post by NothingButWaleed)
bro this was like 2 years ago, ive already done my GCSE exams like 7 months agoPosted on the TSR App. Download from Apple or Google Play -
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- 06-03-2018 18:19
(Original post by Y11_Maths)
How did you do?
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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- 06-03-2018 18:20
(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!Posted on the TSR App. Download from Apple or Google Play
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