Respuesta :

[tex]\mathrm{\mathbf{Vertices}\ \to\ P(3,-1);\ Q(3,2);\ R(-3,-3)}\\\\ \mathrm{S=\dfrac{1}{2}|D|\ \to\ S=\dfrac{1}{2}\left|\begin{array}{ccc}\mathrm{x_P}&\mathrm{y_P}&1\\\mathrm{x_Q}&\mathrm{y_Q}&1\\\mathrm{x_R}&\mathrm{y_R}&1\end{array}\right|\ \to\ S=\dfrac{1}{2} \left|\begin{array}{ccc}3&-1&1\\3&2&1\\-3&-3&1\end{array}\right|\ \to}\\\\\\ \mathrm{\to\ S=\dfrac{1}{2}\big(6+3-9+3+9+3)\ \to\ \boxed{\mathbf{S=\dfrac{15}{2}\ u.a.}}}[/tex]
See, please, the offered decision.
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