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

The area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) can be found using the formula:
$$	ext{Area}=\frac{1}{2} \begin{vmatrix}x_1&y_1&1\x_2&y_2&1\x_3&y_3&1\end{vmatrix}$$
Plugging in the values, we have:
\begin{align*}
	ext{Area}&=\frac{1}{2} \begin{vmatrix} \frac{3}{2} & 0 & 1 \ 2 & 6 & 1 \ 1 & 4 & 1 \end{vmatrix}\
&=\frac{1}{2}\left[\left(\frac{3}{2} \cdot 6 \cdot 1 + 2 \cdot 4 \cdot 1 + 0 \cdot 1 \cdot 1ight) - \left(1 \cdot 6 \cdot 1 + \frac{3}{2} \cdot 4 \cdot 1 + 2 \cdot 0 \cdot 1ight)ight]\
&=\frac{1}{2}\left[9-3ight]\
&=\frac{3}{2}
\end{align*}
Thus, the answer is choice $\boxed{	extbf{(C)}}$.

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

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