Question

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.

Answer 1

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 =
`\left[\begin{array}{ccc} x_(1)&y_(1)&1\\x_(2)&y_(2)&1\\x_(3)&y_(3)&1\end{array}\right]`

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