# Solving a question regarding area of triangle in x-y plane

• Jul 9th 2010, 10:25 PM
Glitch
Solving a question regarding area of triangle in x-y plane
This is the question:

Find the area of the triangle with vertices (1, -2), (-1, 3), (2, 4).

I have already solved this, however my method seems quite lengthy. I just want to know if there's a better way to do it.

My method:
1. Label the vertices a, b and c respectively.
2. Find the distance between the vertices using the distance formula.
3. Let the largest distance be the 'base' of the triangle.
4. Find the gradient of the base, choose one of its verticies, and use the point-gradient formula to get the equation of the line.
5. Find the line which cuts through the base, and passes through the opposite vertex. Do this by getting the perpendicular gradient of the base, then using the opposite vertex, find the equation of the line using the point-gradient formula.
6. Now we simultaneously solve the equation for the base, and the perpendicular line that cuts it. This will give the point at where the line cuts the base.
7. We know where the line cuts the base, so we use that to find the height of the triangle, by finding the distance between the point of intersection and the opposite vertex.
8. Now it's just a matter of 1/2 * base * height

Is there a better way?
• Jul 9th 2010, 11:34 PM
red_dog
If $A(x_1,y_1), \ B(x_2,y_2), \ C(x_3,y_3)$ are the vertices of the triangle ABC, then the area of the triangle ABC is

$A=\frac{1}{2}|\Delta|$

where $\Delta=\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix}$
• Jul 10th 2010, 12:42 AM
Glitch
I'm not sure what you're trying to say.
• Jul 10th 2010, 02:24 AM
mr fantastic
Quote:

Originally Posted by Glitch
I'm not sure what you're trying to say.

If that's the case then it involves ideas (such as the scalar triple product) that you do not yet know.
• Jul 10th 2010, 03:34 AM
Prove It
Just find the distance between each of the vertices to give you the three lengths of the triangle, then apply Heron's Formula.

Heron's Formula: For a triangle of side lengths $a, b, c$

$A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a + b + c}{2}$.
• Jul 10th 2010, 03:57 AM
Glitch
We haven't learnt that, but thanks!
• Jul 10th 2010, 04:16 AM
mr fantastic
Quote:

Originally Posted by Glitch
We haven't learnt that, but thanks!

Unlike the formula in post #2, it requires no special mathematical background. So consider it something you have now learned.
• Jul 10th 2010, 05:54 AM
Glitch
I've memorised the formula. Thanks again!