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 $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ can be found by taking half of the determinant of the matrix
$$\begin{pmatrix}x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1\end{pmatrix}$$
Thus, for this triangle with vertices $\left(\frac{3}{2},0\right)$, $(2,6)$, and $(1,4)$, we have
$$\frac{1}{2}\begin{vmatrix}\frac{3}{2} & 0 & 1 \ 2 & 6 & 1 \ 1 & 4 & 1\end{vmatrix} = \frac{1}{2}\left(\frac{9}{2} - 6\right) = \boxed{\textbf{(C)}\ \frac{5}{2}}.$$

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

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