Answer
$${\rm{9}}{\rm{.5}}\,{\rm{unit}}{{\rm{s}}^2}$$
Work Step by Step
$$\eqalign{
& {\rm{Let\, the\, points }}P\left( {\underbrace 2_{{x_1}},\underbrace 5_{{y_1}}} \right),\,\,Q\left( {\underbrace { - 1}_{{x_2}},\underbrace 3_{{y_2}}} \right),\,\,R\left( {\underbrace 4_{{x_3}},\underbrace 0_{{y_3}}} \right) \cr
& {\rm{Calcule \,the\, area \,using}} \cr
& D = {1 \over 2}\left| {\matrix{
{{x_1}} & {{y_1}} & 1 \cr
{{x_2}} & {{y_2}} & 1 \cr
{{x_3}} & {{y_3}} & 1 \cr
} } \right| \cr
& {\rm{Substitute\, the\, values}} \cr
& 2D = \left| {\matrix{
2 & 5 & 1 \cr
{ - 1} & 3 & 1 \cr
4 & 0 & 1 \cr
} } \right| \cr
& {\rm{Calculate \,the\, determinant }} \cr
& 2D = 4\left| {\matrix{
5 & 1 \cr
3 & 1 \cr
} } \right| - 0\left| {\matrix{
2 & 1 \cr
{ - 1} & 1 \cr
} } \right| + 1\left| {\matrix{
2 & 5 \cr
{ - 1} & 3 \cr
} } \right| \cr
& 2D = 4\left( {5 - 3} \right) - 0 + \left( {6 + 5} \right) \cr
& 2D = 19 \cr
& D = {{19} \over 2} = 9.5 \cr
& {\rm{The\, area\, of\, the\, triangle \,is\, 9}}{\rm{.5}}\,{\rm{unit}}{{\rm{s}}^2} \cr} $$
