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=
`(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=
`(1)/(2)\begin{vmatrix}3&3&1 \\ 3&-1&1\\ -2&-5&1\end{vmatrix}`

Therefore, Area=
`(1)/(2)* [[3(-1* 1-(-5)* 1]-3[3* 1-(-2)* 1]+1[3* -5-2]]= (1)/(2)* -20=-10`

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