# Using integration, find the area of a triangle whose vertices are (1, 0), (2, 2) and (3, 1).

Given: Vertices of the trinagle A(1, 0), B(2, 2) and C(3, 1)

$\\ \text {Equations of } \mathrm{AB} : \\ y-0=\frac{2-0}{2-1}({x}-1)\\\\ y = 2x-2$

$\\ \text {Equations of } \mathrm{AC} : \\ y-0=\frac{1-0}{3-1}({x}-1)\\\\ 2y = x-1$

$\\ \text {Equations of } \mathrm{BC} : \\ y-2=\frac{1-2}{3-2}({x}-2)\\\\ y = 4-x$

$\\ \text {Area of triangle ABC } =2 \int_{1}^{2}(\mathrm{x}-1) \mathrm{d} \mathrm{x}+\int_{2}^{3}(4-\mathrm{x}) \mathrm{d} \mathrm{x}-\frac{1}{2} \int_{1}^{3}(\mathrm{x}-1) \mathrm{d} \mathrm{x} \\\\\ =2\left[\frac{(\mathrm{x}-1)^{2}}{2}\right]_{1}^{2}-\left[\frac{(4-\mathrm{x})^{2}}{2}\right]_{2}^{3}-\frac{1}{2}\left[\frac{(\mathrm{x}-1)^{2}}{2}\right]_{1}^{3} \\\\ =2(\frac{1}{2}-0) - (\frac{1}{2}-2)-\frac{1}{2}(2-0)\\\\ =1+\frac{3}{2}-1\\\\=\frac{3}{2} \ \mathrm{square} \text { units }$

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