Using Integration, find the area of triangle whose vertices are ( $\displaystyle -1,1$ ), $\displaystyle (0,5)$ and $\displaystyle (3,2)$.

Correct Figure Equation of the lines $\displaystyle \mathbf{A B}, \mathbf{B C}$ and $\displaystyle \mathbf{A C}$ are:\begin{aligned} y & =4 x+5 ; y=5-x ; y=\frac{x}{4}+\frac{5}{4} \text { respectively, } \operatorname{ar}(\Delta A B C) & =\int_{-1}^{0}(4 x+5) d x+\int_{0}^{3}(5-x) d x-\frac{1}{4} \int_{-1}^{3}(5+x) d x & \left.\left.\left.=\frac{1}{8}(4 x+5)^{2}\right]_{-1}^{0}-\frac{1}{2}(5-x)^{2}\right]_{0}^{3}-\frac{1}{8}(5+x)^{2}\right]_{-1}^{3} & =3+\frac{21}{2}-6=\frac{15}{2} \end{aligned} $$