Question

What is the area of a triangle whose vertices are D(3,3), E(3,-1), and F(-2,-5)?

Answer 1

There is a formula which employs the use of determinants and which helps us calculate the area of a triangle if the vertices are given as $(x_1,y_1),(x_2,y_2),(x_3,y_3)$. The formula is as shown below:

Area=$\frac{1}{2}\begin{vmatrix}x_1&y_1&1 \\ x_2&y_2&1\\ x_3&y_3&1\end{vmatrix}$

Now, in our case, we have: $(x_1,y_1)=(3,3)$

$(x_2,y_2)=(3,-1)$, and

$(x_3,y_3)=(-2,-5)$

Thus, the area in this case will become:

Area=$\frac{1}{2}\begin{vmatrix}3&3&1 \\ 3&-1&1\\ -2&-5&1\end{vmatrix}$

Therefore, Area=$\frac{1}{2}\times [[3(-1\times 1-(-5)\times 1]-3[3\times 1-(-2)\times 1]+1[3\times -5-2]]= \frac{1}{2}\times -20=-10$

We know that area cannot be negative, so the area of the given triangle is 10 squared units.