Plot each point& connect them to form the triangle. Find the area of the triangle using  2 Formulas A(-5,3);  B(6, 0);  C(5,5) Plot each point of the following points, and connect them to form the triangle ABC.  Find the area of the triangle using two techniques;  Heron's Formula, and the three-point coordinate formula (see Math Open Reference, 2009).  Show your work in detail.  Prove that both methods yield the same results

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You may find the area of triangle using 3x3 determinant, where the first column consists of values of x coordinates of vertices, the second column consists of values of y coordinates of vertices and the third column consists of values equal to 1 such that:

`S = (1/2)|[(-5,3,1),(6,0,1),(5,5,1)]|`

`S = (1/2)(0+30+15-0+25-18) => S = 26`

You may evaluate the area of triangle using Heron's formula, but you need to find the lengths of the sides of triangle first.

`S = (sqrt((a+b+c)(b+c-a)(a+c-b)(a+b-c)))/4`

You may evaluate the lengths of teh sides using distance formula such that:

`d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Reasoning by analogy yields:

`[AB] = sqrt((6+5)^2+(0-3)^2)=>[AB] = sqrt130 = c`

`[AC] = sqrt((5+5)^2+(5-3)^2) => [AC] = sqrt104 = b`

`[BC] = sqrt((5-6)^2+(5-0)^2) = >[BC] = sqrt 26 = a`

`S = (sqrt((sqrt 26+sqrt104+sqrt130)(sqrt104+sqrt130-sqrt 26)(sqrt 26+sqrt130-sqrt104)(sqrt 26+sqrt104-sqrt130)))/4`

`S = (sqrt((sqrt 26+sqrt104)^2-130)(130-(sqrt 26-sqrt104)^2))/4`

`S = (sqrt(130(sqrt 26+sqrt104)^2 - (26-104)^2 - 130^2 + 130(sqrt 26+sqrt104)^2))/4`

`S = sqrt(260(sqrt 26+sqrt104)^2 -22984)/4 => S = 26`

Hence, evaluating the area of triangle using both formulas yields `S = 26` .

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