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If A(-5,7), B(-4,-5), C(-1,-6) and D(4,5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

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After plotting the quadrilateral in a Cartesian plane, you can see that it is not a particular quadrilateral. Hence, you need to divide it into two triangles. Let's take ABC and ADC.

The area of a triangle with vertices known is  given by the matrix
M = [tex] \left[\begin{array}{ccc} x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right] [/tex]

Area = 1/2· | det(M) |
        = 1/2· | x₁·y₂ - x₂·y₁ + x₂·y₃ - x₃·y₂ + x₃·y₁ - x₁·y₃ |
        = 1/2· | x₁·(y₂ - y₃) + x₂·(y₃ - y₁) + x₃·(y₁ - y₂) |

Therefore, the area of ABC will be:
A(ABC) = 1/2· | (-5)·(-5 - (-6)) + (-4)·(-6 - 7) + (-1)·(7 - (-5)) |
             = 1/2· | -5·(1) - 4·(-13) - 1·(12) |
             = 1/2 | 35 |
             = 35/2

Similarly, the area of ADC will be:
A(ABC) = 1/2· | (-5)·(5 - (-6)) + (4)·(-6 - 7) + (-1)·(7 - 5) |
             = 1/2· | -5·(11) + 4·(-13) - 1·(2) |
             = 1/2 | -109 |
             = 109/2

The total area of the quadrilateral will be the sum of the areas of the two triangles:

A(ABCD) = A(ABC) + A(ADC) 
               = 35/2 + 109/2
               = 72

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