Answer
\[{\text{The area of the triangle is 1uni}}{{\text{t}}^2}\]
Work Step by Step
\[\begin{gathered}
{\text{Let the points }}P\left( {\underbrace 0_{{x_1}},\underbrace 0_{{y_1}}} \right),\,\,Q\left( {\underbrace 0_{{x_2}},\underbrace 2_{{y_2}}} \right),\,\,R\left( {\underbrace 1_{{x_3}},\underbrace 4_{{y_3}}} \right) \hfill \\
{\text{Calcule the area using}} \hfill \\
D = \frac{1}{2}\left| {\begin{array}{*{20}{c}}
{{x_1}}&{{y_1}}&1 \\
{{x_2}}&{{y_2}}&1 \\
{{x_3}}&{{y_3}}&1
\end{array}} \right| \hfill \\
{\text{Substitute the values}} \hfill \\
2D = \left| {\begin{array}{*{20}{c}}
0&0&1 \\
0&2&1 \\
1&4&1
\end{array}} \right| \hfill \\
{\text{Calculate the determinant}} \hfill \\
2D = 0\left| {\begin{array}{*{20}{c}}
2&1 \\
4&1
\end{array}} \right| - 0\left| {\begin{array}{*{20}{c}}
0&1 \\
1&1
\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}
0&2 \\
1&4
\end{array}} \right| \hfill \\
2D = \left( 0 \right)\left( 4 \right) - \left( 2 \right)\left( 1 \right) \hfill \\
2D = - 2 \hfill \\
D = - 1 \hfill \\
{\text{Calculate the absolute value of }}D \hfill \\
D = 1 \hfill \\
{\text{The area of the triangle is 1uni}}{{\text{t}}^2} \hfill \\
\end{gathered} \]
