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# Maths C1 Coordinate Geometry

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1. I'm stuck on a question that I remember doing but I've ended up forgetting.

Coordinates given for ABCD (-5,0),(0,5),(3,4)(4,-3)

Lines AC and BD intersect at E.
> Show that the coordinates of E are (1,3)

How do I work out the intersection point?

(They're perpendicular)

Equation AC is x-2y+5 = 0
Equation BD is y=-2x+5

2. (Original post by slot_)
I'm stuck on a question that I remember doing but I've ended up forgetting.

Coordinates given for ABCD (-5,0),(0,5),(3,4)(4,-3)

Lines AC and BD intersect at E.
> Show that the coordinates of E are (1,3)

How do I work out the intersection point?

(They're perpendicular)

Equation AC is x-2y+5 = 0
Equation BD is y=-2x+5

If you draw two lines on a graph and make them intersect, you will see that their intersection has the same co-ordinates, which means the same point is on the two different lines and so satisfies both equations.

Does that give you a hint as to how you can solve those two equations at the same time?
3. (Original post by SeanFM)
If you draw two lines on a graph and make them intersect, you will see that their intersection has the same co-ordinates.

Does that give you a hint as to how you can solve those two equations at the same time?
Simultaneous?
4. (Original post by SeanFM)
If you draw two lines on a graph and make them intersect, you will see that their intersection has the same co-ordinates.

Does that give you a hint as to how you can solve those two equations at the same time?
Hmmmmmmmmmmmmmmmmmmm..........

Same time, eh?........

Yeh I don't get it :/
5. (Original post by slot_)
Simultaneous?
Precisely
6. (Original post by SeanFM)
Precisely
Derp, I got it now. Just needed the pointer lol

For anyone who may want to correct me

AC Gradient = 1/2, Equation = 2y = x+5.
BD = 2y = -2x + 5 with Gradient of -2

-2x + 5 = 1/2x + 5/2
when you multiply by 2 to get whole numbers
-4x + 10 = x + 5
rearrange
(-5) + 10 = x + 4x
5 = 5x
x = 1
Substitute in to later equation so 2y = x + 5
2y = 1 + 5
2y = 6
y = 6/3
y = 3

(1,3)
7. (Original post by slot_)
Derp, I got it now. Just needed the pointer lol

For anyone who may want to correct me

AC Gradient = 1/2, Equation = 2y = x+5.
BD = 2y = -2x + 5 with Gradient of -2

-2x + 5 = 1/2x + 5/2
when you multiply by 2 to get whole numbers
-4x + 10 = x + 5
rearrange
(-5) + 10 = x + 4x
5 = 5x
x = 1
Substitute in to later equation so 2y = x + 5
2y = 1 + 5
2y = 6
y = 6/3
y = 3

(1,3)
A good way to check your answer is to see if the point satisfies both equations.. after all, that is what you're finding
8. (Original post by SeanFM)
Precisely
OKAY I THINK I GOT IT

Step 1: Rearrange equations

Step 2: Represent in matrix form

Step 3: Find the inverse matrix

Step 4: Pre-multiply by the inverse matrix

Step 5: Clean up

So:

x=1
y=3

DID I GET IT RIGHT?!!?!?!
9. (Original post by slot_)
Derp, I got it now. Just needed the pointer lol

For anyone who may want to correct me

AC Gradient = 1/2, Equation = 2y = x+5.
BD = 2y = -2x + 5 with Gradient of -2

-2x + 5 = 1/2x + 5/2
when you multiply by 2 to get whole numbers
-4x + 10 = x + 5
rearrange
(-5) + 10 = x + 4x
5 = 5x
x = 1
Substitute in to later equation so 2y = x + 5
2y = 1 + 5
2y = 6
y = 6/3
y = 3

(1,3)
ur methud is confusing :s
10. (Original post by RDKGames)
OKAY I THINK I GOT IT

Step 1: Rearrange equations

Step 2: Represent in matrix form

Step 3: Find the inverse matrix

Step 4: Pre-multiply by the inverse matrix

Step 5: Clean up

So:

x=1
y=3

DID I GET IT RIGHT?!!?!?!
Correct answer but minus swag points for not using gaussian elimination. Sorry
11. (Original post by seanfm)
correct answer but minus swag points for not using gaussian elimination. Sorry
this is why i'll fail c1
12. (Original post by SeanFM)
Correct answer but minus swag points for not using gaussian elimination. Sorry
On a serious note, do you reckon they would actually award marks for this method in C1???
13. (Original post by RDKGames)
On a serious note, do you reckon they would actually award marks for this method in C1???
I'd hope so. May require some consultation with a senior exam officer or something, and some investigation into how and why the student is using such a method. Fancy but time consuming and superfluous
14. solve the simultaneous equations then BAM you get x coordinate and y coordinate

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