# 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...

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

*print*Print*list*Cite

### 1 Answer

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` .**