Maths C1 Coordinate Geometry

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?

Notes: AC Gradient = 1/2
BD Gradient = -2/1
(They're perpendicular)

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

Please help

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?

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?

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)

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)

OKAY I THINK I GOT IT

Step 1: Rearrange equations
$x-2y=-5$
$2x+y=5$

Step 2: Represent in matrix form

$\begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -5 \\ 5 \end{bmatrix}$

Step 3: Find the inverse matrix

$\begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}^{-1} = \frac{1}{5}\begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}$

Step 4: Pre-multiply by the inverse matrix
$\begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}^{-1} \begin{bmatrix} -5 \\ 5 \end{bmatrix}$

$\Rightarrow \mathbb{I} \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} -5 \\ 5 \end{bmatrix}$

Step 5: Clean up
$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 5 \\ 15 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$

So:

x=1
y=3


DID I GET IT RIGHT?!!?!?!

correct answer but minus swag points for not using gaussian elimination. Sorry

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???