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#

Given 3 vertices of a triangle #(x_1,y_1)#, #(x_2,y_2)#, and #(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#