Answer:
![\boxed{\mathsf{A} \triangle = \red{(67)/(2)u.a}}](https://img.qammunity.org/2021/formulas/mathematics/high-school/f006b548gjo5765j77javqc13jt8ucft4b.png)
Explanation:
Let's follow up with the solution. Considering a triangle with the vertices
,
and
, have a look at the representation in the cartesian plan.
From this representation we can say that the area (A) of a triangle through the knowledge of analytical geometry is given by the determinant of the vertices divided by two, mathematically,
![\mathsf{A} \triangle = \frac{\left| \begin{array}{ccc} \mathsf{x_A} & \mathsf{y_A }& 1 \\ \mathsf{x_B} & \mathsf{ y_B} & 1 \\ \mathsf{ x_C} & \mathsf{ y_C} & 1 \end{array} \right|}{2}](https://img.qammunity.org/2021/formulas/mathematics/high-school/hrt306o7dtcfhp0qxl86rfftyjrrnsd5ew.png)
So, applying this knowledge we're going to have,
![\mathsf{A} \triangle = \frac{\left| \begin{array}{ccc} 3 & -7 & 1 \\ 6 & 4 & 1 \\ -2 & -3 & 1 \end{array} \right|}{2}](https://img.qammunity.org/2021/formulas/mathematics/high-school/b4ramem9slwua4sxptnq678099ggyjpj3b.png)
![\mathsf{A} \triangle = (1)/(2)\left[ \left.\begin{array}{ccc} 3 & -7 & 1 \\ 6 & 4 & 1 \\ -2& -3 & 1 \end{array} \right| \begin{array}{cc} 3 & -7 \\ 6 & 4 \\ -2 & -3 \end{array} \right]](https://img.qammunity.org/2021/formulas/mathematics/high-school/ifpdesm7xd37yh76mqpvfw3zsnzaufx7tz.png)
![\mathsf{A} \triangle = (12 + 14 - 18 - (-8 - 9 - 42))/(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/ay7p7svwt0ub9ayl6vwp45og5gsgv4dbjq.png)
![\red{\mathsf{A} \triangle = (67)/(2) = 33,5u.a}](https://img.qammunity.org/2021/formulas/mathematics/high-school/itbjyh4tjpkiae9z4oqd2caj8qsm0re98r.png)
Hope you enjoy it, see ya!)
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DavidJunior17