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(A) Let the vertices of the quadrilateral be $A(-1, -1), B(-4, -2), C(-5, -4),$ and $D(-2, -3)$.To find the area,we divide the quadrilateral into two

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$5$

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(A) Let the vertices of the quadrilateral be $A(-1, -1), B(-4, -2), C(-5, -4),$ and $D(-2, -3)$.To find the area,we divide the quadrilateral into two triangles,$	riangle ABC$ and $	riangle ADC$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area of $	riangle ABC = \frac{1}{2} |(-1)(-2 - (-4)) + (-4)(-4 - (-1)) + (-5)(-1 - (-2))|$$= \frac{1}{2} |(-1)(2) + (-4)(-3) + (-5)(1)| = \frac{1}{2} |-2 + 12 - 5| = \frac{1}{2} |5| = 2.5$.Area of $	riangle ADC = \frac{1}{2} |(-1)(-4 - (-3)) + (-5)(-3 - (-1)) + (-2)(-1 - (-4))|$$= \frac{1}{2} |(-1)(-1) + (-5)(-2) + (-2)(3)| = \frac{1}{2} |1 + 10 - 6| = \frac{1}{2} |5| = 2.5$.Total Area = Area of $	riangle ABC +$ Area of $	riangle ADC = 2.5 + 2.5 = 5$ square units.

Find the area of a quadrilateral having the vertices $(-1, -1), (-4, -2), (-5, -4),$ and $(-2, -3)$.

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