# Math Calculate the area of the triangle ABC with vertices A(2,3), B(1-6), C(0,3).

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The triangle has vertexes A(2,3), B(1, -6), C(0,3)

AB = sqrt( 1 + 81) = sqrt 82

BC = sqrt(1 + 81) = sqrt 82

AC = 2

The area of a triangle is given by the Heron's formula as sqrt [s(s - a)(s - b)(s - c)] where a, b and c are the sides and s is the semi-perimeter.

s = sqrt 82 + 1

sqrt [s(s - a)(s - b)(s - c)]

=> sqrt [(sqrt 82 + 1)(1)(1)( sqrt 82 - 1)]

=> sqrt [ (82 - 1)]

=> 9

**The area of the triangle is 9 square units.**

We'll apply the formula:

S = (1/2)*|det M|

|xA yA 1|

det M = |xB yB 1|

|xC yC 1|

det M = xA*yB + xB*yC + xC*yA - xCyB - xA*yC - xB*yA

We'll replace the coordinates of the vertices A,B,C by their values:

det M = 2*(-6) + 1*3 + 0*3 - 0*(-6) - 2*3 - 1*3

det M = -12 + 3 - 6 - 3

det M = -18

The area of the triangle is:

S = (1/2)*|-18|

**S = 9 square units.**