# Plot each point& connect them to form the triangle. Find the area of the triangle using  2 Formulas A(4,-3);  B(0,-3);  C(4,2) 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

Since the problem provides the coordinates of vertices of triangle, you should use the formula that implies the following determinant  to evaluate the area of triangle such that:

`S = (1/2)|[(x_A,y_A,1),(x_B,y_B,1),(x_C,y_C,1)]|`

`S = (1/2)|[(4,-3,1),(0,-3,1),(4,2,1)]|` = `(1/2)|(-12 + 0 - 12 - (-12) - 8 - 0)|`

`S = (1/2)|-20| =...

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Since the problem provides the coordinates of vertices of triangle, you should use the formula that implies the following determinant  to evaluate the area of triangle such that:

`S = (1/2)|[(x_A,y_A,1),(x_B,y_B,1),(x_C,y_C,1)]|`

`S = (1/2)|[(4,-3,1),(0,-3,1),(4,2,1)]|` = `(1/2)|(-12 + 0 - 12 - (-12) - 8 - 0)|`

`S = (1/2)|-20| = 10`

You may also use Heron's formula but you need to find the lengths of the sides of triangle first, since Heron's formula uses these values such that:

`S = (1/4)sqrt((a+b+c)(a+b-c)(b+c-a)(c+a-b))`

You need to find a,b,c using distance formula such that:

`a = sqrt((x_C-x_B)^2 + (y_C - y_B)^2)`

`a = sqrt((4-0)^2 + (2+3)^2) => a = sqrt(16 + 25) =>a = sqrt41`

`b = sqrt((x_C-x_A)^2 + (y_C - y_A)^2)`

`b = sqrt((4-4)^2 + (2+3)^2) => b = sqrt25 = 5`

`c = sqrt((x_A-x_B)^2 + (y_A - y_B)^2)`

`c = sqrt(16 + 0) => c = 4`

`S = (1/4)sqrt((sqrt41+5+4)(sqrt41+5-4)(5+4-sqrt41)(4+sqrt41-5))`

`S = (1/4)sqrt((9^2 - 41)(41 - 1)) => S = (1/4)sqrt((81-41)(40))`

`S = (1/4)sqrt(40^2) => S = 40/4 => S = 10`

Hence, evaluating the area of triangle using both formulas yields `S = 10` .

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