What is the area of a triangle whose vertices are GC-1, 2), H(5, 2), and K(8, 3)?

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Answer 1 · 2017-06-06T18:38:47

$Area = 3$ $"Area" = 3$

Given 3 vertices of a triangle $(x_{1}, y_{1})$ $(x_1,y_1)$ , $(x_{2}, y_{2})$ $(x_2,y_2)$ , and $(x_{3}, y_{3})$ $(x_3,y_3)$

This reference, Applications of Matrices and Determinants tells us how to find the area:

$"Area" = +-1/2| (x_1,y_1,1), (x_2,y_2,1), (x_3,y_3,1) |$

Using the points $(-1, 2), (5, 2), and (8, 3)$ :

$"Area" = +-1/2| (-1,2,1), (5,2,1), (8,3,1) |$

I use the Rule of Sarrus to compute the value of a $3xx3$ determinant:

$| (-1,2,1,-1,2), (5,2,1,5,2), (8,3,1,8,3) | =$

$(-1)(2)(1)-(-1)(1)(3) + (2)(1)(8)-(2)(5)(1)+(1)(5)(3)-(1)(2)(8) = 6$

Multiply by $1/2$ :

$"Area" = 3$