The figure of quadrilateral whose vertices are A(– 4, 5), B(0, 7), C(5, –5) and D(–4, –2) is shown in the below figure. Area of the quadrilateral ABCD = area of ΔABC + area of ΔADC ... (i) Now, Area of ΔABC = $\frac{1}{2}\begin{vmatrix} -4 & 5 & 1 \\ 0 & 7 & 1 \\ 5 & -5 & 1 \end{vmatrix}$ = $\frac{1}{2}$ |- 4 (7 + 5) - 0 (5 + 5) + 5 (5 - 7)| = $\frac{1}{2}$ |- 4 (12) + 5 (- 2)| = $\frac{1}{2}$ |- 58| = 29 sq. units Also, Area of ΔADC = $\frac{1}{2}\begin{vmatrix} -4 & 5 & 1 \\ -4 & -2 & 1 \\ 5 & -5 & 1 \end{vmatrix}$ = $\frac{1}{2}$ |63| = $\frac{63}{2}$ sq.units ∴ Area of quadrilateral ABCD = 29 + $\frac{63}{2}$ = $\frac{121}{2}$ sq. units [From (i)]