C1 areas

How would you work out the area of a triangle using coordinates?

You can use the shoelace formula: https://en.m.wikipedia.org/wiki/Shoelace_formula

Ah Jesus your kidding me. That's too long

Ah Jesus your kidding me. That's too long

Lol well that's a general formula...

For your situation, you have a triangle so $n=3$ hence the formula truncates down to

$A=\frac{1}{2}(x_1y_2+x_2y_3+x_3y_1 - x_2y_1-x_3y_2-x_1y_3)$

As I alluded to in my post above, you don't really need the algebraic form of the formula. You can just write a list of the coordinates of each point in order, going down the page, and then after you the last point, write down the first point again. Then draw diagonal lines between the numbers going down/right and down/left. Multiply along the diagonals and add the products from the down/right diagonals, but subtract the products from the down/left diagonals. Finally, take the absolute value of the result, then halve the answer to get the area.

How would you work out the area of a triangle using coordinates?

I don't see a nice A-Level way to go about it that would put alternative methods outside A-Level to shame.

If you want to do the 1/2*base*height thing then it's cumbersome since you need to determine a 4th point such that your triangle is split into 2 right-angled ones, and apply your usual formula to both before adding them.


TBH, you haven't specified what points you got to work with. It could very well be the case where they already form a right-angled triangle so this is very easy! At the moment you're asking a general question which has general answers.

It's the question from the c1 edexcel January 2012 paper question 6. It's 2 right angles stuck together.

Well then that's easy...

Total area is just the area of triangle COB + the area of triangle AOB.

These are both right angled triangles. You can find the lengths of their sides. For example, length OB is just the y-coordinate of B.

If the coordinates where not on the same x axis would you use pythagoras to get the distances?

Couldn't you have just worked out the length of OB and multiplied it by AC and divided by 2 to get the triangle? Or do you have to split it into 2 right angles?