Question

calculate the area of the triangle with the following vertices (3, -7), (6, 4), (-2, -3)

Answer 1

Answer:

$\boxed{\mathsf{A} \triangle = \red{\dfrac{67}{2}u.a}}$

Step-by-step explanation:

Let's follow up with the solution. Considering a triangle with the vertices $\mathsf{A(x_A, y_A)}$, $\mathsf{B(x_B, y_B)}$ and $\mathsf{C(x_C, y_C)}$, 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,

$\mathsf{A} \triangle = \dfrac{\left| \begin{array}{ccc} \mathsf{x_A} & \mathsf{y_A }& 1 \\ \mathsf{x_B} & \mathsf{ y_B} & 1 \\ \mathsf{ x_C} & \mathsf{ y_C} & 1 \end{array} \right|}{2}$

So, applying this knowledge we're going to have,

$\mathsf{A} \triangle = \dfrac{\left| \begin{array}{ccc} 3 & -7 & 1 \\ 6 & 4 & 1 \\ -2 & -3 & 1 \end{array} \right|}{2}$

$\mathsf{A} \triangle = \dfrac{1}{2}\left[ \left.\begin{array}{ccc} 3 & -7 & 1 \\ 6 & 4 & 1 \\ -2& -3 & 1 \end{array} \right| \begin{array}{cc} 3 & -7 \\ 6 & 4 \\ -2 & -3 \end{array} \right]$

$\mathsf{A} \triangle = \dfrac{12 + 14 - 18 - (-8 - 9 - 42)}{2}$

$\red{\mathsf{A} \triangle = \dfrac{67}{2} = 33,5u.a}$

Hope you enjoy it, see ya!)

$\green{\mathsf{FROM}}:$ Mozambique, Maputo – Matola City – T-3

DavidJunior17