Question

Find the area of the triangle with vertices $(2,3)$, $(0,1)$, and $(1,2)$

• A. $\frac{1}{2}$ square unit
• B. $0$ square unit
• C. $2$ square unit
• D. $2\frac{1}{2}$ square unit

Answer 1

The correct option is B

$0$ square unit

Finding the area of triangle:

Let the points be, $A(2,3)$, $B(0,1)$ and$C(1,2)$, then

Area of the triangle in determinant form is given by

$\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Here,

$x_1=2,x_2=0,x_3=1,y_1=3,y_2=1,y_3=2$

putting these values, we get

$$\begin{gathered} =\frac{1}{2}\left|\begin{array}{ccc}2& 3& 1\\ 0& 1& 1\\ 1& 2& 1\end{array}\right| \\ =\frac{1}{2}\left[2\right(1-2)-3(-1)+1(-1\left)\right] \\ =\frac{1}{2}(-2+3-1) \\ =\frac{1}{2}(3-3) \\ =\frac{1}{2}\left(0\right) \\ =0\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$

Hence the correct option is (B).