A triangle has vertices $\left( \frac { 3 } { 2 } , 0 \right)$ , $( 2,6 )$, and $( 1,4 )$.
Use a determinant to find its area
A) $\frac { 4 } { 3 }$

B) $\frac { 5 } { 3 }$

C) $\frac { 5 } { 2 }$

D) $\frac { 2 } { 5 }$

E) $2$

Question

A triangle has vertices $\left( \frac { 3 } { 2 } , 0 \right)$ , $( 2,6 )$, and $( 1,4 )$.
 Use a determinant to find its area
A) $\frac { 4 } { 3 }$

B) $\frac { 5 } { 3 }$

C) $\frac { 5 } { 2 }$

D) $\frac { 2 } { 5 }$

E) $2$

Quizplus · Accepted Answer

We can use the determinant formula to find the area of the triangle. The determinant of the coordinates of the vertices is:
$$\begin{vmatrix}
\frac{3}{2} & 0 & 1\ 
2 & 6 & 1\ 
1 & 4 & 1
\end{vmatrix} = \frac{3}{2}\begin{vmatrix}
6 & 1\ 
4 & 1
\end{vmatrix} - 2\begin{vmatrix}
2 & 1\ 
1 & 1
\end{vmatrix}+ \frac{1}{2}\begin{vmatrix}
2 & 6\ 
1 & 4
\end{vmatrix}$$
Simplifying the determinants, we get:
$$\begin{vmatrix}
\frac{3}{2} & 0 & 1\ 
2 & 6 & 1\ 
1 & 4 & 1
\end{vmatrix} = \frac{5}{2}$$
Therefore, the area of the triangle is $\frac{1}{2}$ times the absolute value of the determinant, which is $\frac{5}{4}$. The closest answer choice to this value is $\boxed{\textbf{(C) }\frac{5}{2}}$.

A Triangle Has Vertices (32,0)\left( \frac { 3 } { 2 } , 0 \right)(23​,0)

A Triangle Has Vertices $\left( \frac { 3 } { 2 } , 0 \right)$