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# C2. Circles/ Coordinate geometry... q watch

1. Hi
Can someone pls tell me of a better way to solve this q?
2. Can you send the actual question?

I'd consider Heron's formula to solve it in a few lines.
3. (Original post by I-try)
Hi
Can someone pls tell me of a better way to solve this q?
This is the standard way though. All you do is determine the coordinates of C by using the fact that A and C have (2,-1) as the midpoint, then just find the lengths of AB and CB and use the formula for the area of a right-angled triangle.
This is all that the solution shows.

The solution seems pretty long-winded and some shortcuts can be taken, such as recognising the coordinate of C as (1,-9) straight away by recognising the vector from A to the centre is (-1,-8) so just add that onto the centre coordinate. Then the lengths of the legs are just punching numbers into your calc
4. Never heard of that formula
5. (Original post by RDKGames)
This is the standard way though. All you do is determine the coordinates of C by using the fact that A and C have (2,-1) as the midpoint, then just find the lengths of AB and CB and use the formula for the area of a right-angled triangle.
This is all that the solution shows.
But isn't there a better way to find the midpoint? Like they've found it by plotting the triangle on a graph right?
6. (Original post by I-try)
Never heard of that formula
You can google and learn about Heron's formula. It's mostly used in Maths competitions and Olympiads however you can use it in A level Maths (and get the marks). It's a way of calculating area of a triangle provided three sides.

It's very easy to get the length of 3 sides and then just plug it into his formula. Also, it can be used in C4 vectors areas/volumes problems so I'd recommend giving it a read.
7. (Original post by I-try)
But isn't there a better way to find the midpoint? Like they've found it by plotting the triangle on a graph right?
You're practically told what the midpoint. It says the circle has center (2,-1) and the points A,C are two ends of the diameter, and we know that the mid point of the diameter must be the center for every circle.
8. Oh never mind I get it, I get it. Sorry and thank you both! I just didn't understand the second step in the solution bank >< so I asked for a better way to solve the q
9. (Original post by thekidwhogames)
I'd consider Heron's formula to solve it in a few lines.
This is more work than the usual method. With Heron's, you need to work out what C is and calculate all three side lengths. Whereas in the usual method you just work out C and calculate two side lengths.

No point sledgehammering this problem with it.
10. (Original post by RDKGames)
This is more work than the usual method. With Heron's, you need to work out what C is and calculate all three side lengths. Whereas in the usual method you just work out C and calculate two side lengths.

No point sledgehammering this problem with it.
I suppose that's true for this problem but I think OP should still read up on Heron's as his formula can make other questions (and C4 vectors) much easier. For instance, in some coordinate geometry questions you'd have to use the cosine rule and apply 1/2 abSinC or even split the non-right triangle into separate shapes etc - in these cases, Heron's makes it much easier (it's also very quick to calculate lengths; you just apply the formula into a calculator (takes maybe 15 seconds)).
11. (Original post by thekidwhogames)
I suppose that's true for this problem but I think OP should still read up on Heron's as his formula can make other questions (and C4 vectors) much easier. For instance, in some coordinate geometry questions you'd have to use the cosine rule and apply 1/2 abSinC or even split the non-right triangle into separate shapes etc - in these cases, Heron's makes it much easier (it's also very quick to calculate lengths; you just apply the formula into a calculator (takes maybe 15 seconds)).
Will do. Thank you ^-^

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