Finding the area of a triangle, given three coordinates?

Question

I have been attempting the following question:

I can get the answer to part (a) and part (b) but am struggling to find the area of the triangle from the three coordinates. Here is my working:

I have tried showing he triangle is right-angled, and have used Heron’s formula, but I just can’t seem to get the correct answer. Any help would be greatly appreciated!

Edit: I have now applied the shoelace method (pictured below) and have come to the correct answer:

Area $ = \text{abs(((x-coord of Q) - (x-coord of R))} \times \text{y-coord of P)}$ — ab123, Sep 08 '19 at 08:05
The site Area of the Triangle Formed by Three co-ordinate Points explains and then gives an explicit formula for what you're asking about. — John Omielan, Sep 08 '19 at 08:07
Since point $Q,R$ lie on the x-axis, you get the base of the triangle. By knowing the vertical distance from $P$ to segment $QR$, you should get the height of the triangle. — Mythomorphic, Sep 08 '19 at 08:13
I have just used the shoelace formula to get the correct answer, and it seems like quite a good general method. I am not sure where I went wrong in my original working, but I have learnt a new method now. — Jamminermit, Sep 08 '19 at 08:16
@Mythomorphic And height of the triangle is simply the absolute value of y-coordinate of $P$ — ab123, Sep 08 '19 at 08:17
@ab123 In your first comment, the area of the triangle would be $\frac{1}{2}$ of what you stated as the formula is $\frac{1}{2}bh$ where $b$ is the base and $h$ is the height. — John Omielan, Sep 08 '19 at 08:19
Do you know the relationship of determinants and/or cross products to areas of triangle and paralellograms? For a triangle, the shoelace formula is precisely the determinant of a certain $3\times3$ matrix. — amd, Sep 09 '19 at 01:09
@amd When I saw the shoelace formula I recognised it as the determinant, but I wasn’t sure why the determinant would yield the area. Also, I didn’t use a 3x3 matrix, it was 2x4. — Jamminermit, Sep 09 '19 at 05:16

score 1 · Answer 1 · answered Sep 09 '19 at 05:25

1

Once you’ve got the vertices of the triangle, its area can be computed via a determinant: $$\frac12 \begin{vmatrix}P-R\\Q-R\end{vmatrix} = \frac12\begin{vmatrix}x_P&y_P & 1\\x_Q&y_Q & 1\\x_R&y_R & 1\end{vmatrix}.$$ You can view the first determinant as $\lvert(P-R)\times(Q-R)\rvert$, although strictly speaking the cross product isn’t defined in $\mathbb R^2$.

answered Sep 09 '19 at 05:25

amd

53,693

I have edited my question with the shoelace formula that I am familiar with attached. I am fairly new the matrices, so I just assumed that the method pictured was taking the determinant – Jamminermit Sep 09 '19 at 05:55

Finding the area of a triangle, given three coordinates?

1 Answers1