Find the area of the triangle with vertices at

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You need to use the following formula to evaluate the area of triangle whose vertices are provided by the problem, such that:

`A = (1/2)|Delta|`

`Delta = [(x_A,y_A,z_A,1),(x_B,y_B,z_B,1),(x_C,y_C,z_C,1),(1,1,1,1)]`

`Delta = [(-3,4,0,1),(-4,6,-5,1),(-1,1,-5,1),(1,1,1,1)]`

You need to evaluate the determinant using the first line (it contains a zero) such that:

`Delta = (-1)^2*(-3)*[(6,-5,1),(1,-5,1),(1,1,1)] + (-1)^3*4*[(-4,-5,1),(-1,-5,1),(1,1,1)] + (-1)^5*1*[(-4,6,-5),(-1,1,-5),(1,1,1)]`

`Delta = -3(-30 + 1 - 5 + 5 - 6 + 5) - 4(20 - 1 - 5 + 5 + 4 - 5) - (-4 + 5 - 30 + 5 - 20 + 6)`

`Delta = 90 - 72 + 38 = 56`

`A = 56/2 => A = 28`

**Hence, evaluating the areaof triangle Delta ABC yields `A = 28` .**

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