# Right angle Core 1 AQA Maths Question. What is K?!Watch

#1
So doing the maths core 1 AQA exam today, one of the questions showed two points: A(-2,8) and B(5,-6). Point C had coordinates (K, K+1), such that angle BCA is a right angle. Find k. Correct me if any of the question is wrong, but! I have attacked this a few times and it is unanswerable.
0
2 years ago
#2
You had to understand that ACB was 90 and therefore AC would be perpendicular to BC. So you use the gradient equation equation to find an algebraic fraction for both lines. We know that for perpendicular lines M1 x M2 = -1 So you'd sub in the two algebraic fractions you got for AC(M1) and BC(M2). Solve the presented algebraic fraction for k, iassume you did this in GCSE maths
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#3
But to get k, the x and y values had to be 1 apart, hence (k, k+1). And the possible points had to be horizontally and vertically in line with points A and B. This makes c, (-2,-6) or (5,8) which aren't one apart
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2 years ago
#4
A (-2,5) B (8,-6) C (k, k+1) A and B form 2 lines with C, AC and AB. AC and AB intersect at right angles, as stated in the question. Therefore the product of both gradients = -1 (perpendicular rule) gradient AC = (k-4) / (k+2) & BC = (k+7) / (k-8) AC x BC = -1; (k-4)(k+7) / (k+2)(k-8) = -1
Multiply by (k+2)(k-8) to get k^2+3k-28 = -(k^2-6ki-16) Solve to find k.
There was really only one approach to this. Idk how you thought the values had to be 1 apart.
0
2 years ago
#5
I got 9 or -12
0
2 years ago
#6
Got 11/2 and -4
0
2 years ago
#7
(Original post by Xenonical)
A (-2,5) B (8,-6) C (k, k+1) A and B form 2 lines with C, AC and AB. AC and AB intersect at right angles, as stated in the question. Therefore the product of both gradients = -1 (perpendicular rule) gradient AC = (k-4) / (k+2) & BC = (k+7) / (k-8) AC x BC = -1; (k-4)(k+7) / (k+2)(k-8) = -1
Multiply by (k+2)(k-8) to get k^2+3k-28 = -k^2-6ki-16 Solve to find k.
There was really only one approach to this. Idk how you thought the values had to be 1 apart.
dude you should put a bracket before the quadratic and the minus as the whole thing is minus.

If you already intended to do that then lol, just don't wanna create confusion to others who see it.
0
2 years ago
#8

This picture gives a nice geometric description of the problem. Note that the line segment joining A and B is the diameter of the circle (by construction).
0
2 years ago
#9
i think there were several different approaches you couldve done; i dont remember exactly what my answer was but i did it using pythagoras when most others i asked did the gradient thing but i think we got the same answers anyway so its cool

(Original post by Sid.S)
Got 11/2 and -4
yeah these answers ring a bell; 99% sure that's what i got too
0
#10
Thanks for the answers, it makes sense now, i was thinking that they had to be at those two points. I think i got the rest of the paper though so we're all good.
0
2 years ago
#11
(Original post by TomBarton)
So doing the maths core 1 AQA exam today, one of the questions showed two points: A(-2,8) and B(5,-6). Point C had coordinates (K, K+1), such that angle BCA is a right angle. Find k. Correct me if any of the question is wrong, but! I have attacked this a few times and it is unanswerable.
I didn't know how to do this question either. It totally threw me!
0
2 years ago
#12
OMG it makes sense now xD I feel so silly. God I wish I knew this...

(Original post by Xenonical)
A (-2,5) B (8,-6) C (k, k+1) A and B form 2 lines with C, AC and AB. AC and AB intersect at right angles, as stated in the question. Therefore the product of both gradients = -1 (perpendicular rule) gradient AC = (k-4) / (k+2) & BC = (k+7) / (k-8) AC x BC = -1; (k-4)(k+7) / (k+2)(k-8) = -1
Multiply by (k+2)(k-8) to get k^2+3k-28 = -(k^2-6ki-16) Solve to find k.
There was really only one approach to this. Idk how you thought the values had to be 1 apart.
0
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