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

$\frac{15}{2} \text{ units}^2$

The correct answer is Option (1) → $\frac{15}{2} \text{ units}^2$

$(-1, 1)$, $(0, 5)$ and $(3, 2)$

Eq. of side AB: $y - 1 = \frac{5 - 1}{0 + 1}(x + 1)$

$⇒y = 4x + 5$

Eq. of side BC: $y - 5 = \frac{2 - 5}{3 - 0}(x - 0)$

$⇒y = 5 - x$

Eq. of side AC: $y - 1 = \frac{2 - 1}{3 + 1}(x + 1)$

$⇒y = \frac{x + 5}{4}$

$\text{Required area of } \triangle ABC$

$= \int\limits_{-1}^{0} (4x + 5) \, dx + \int\limits_{0}^{3} (5 - x) \, dx - \int\limits_{-1}^{3} \left( \frac{x + 5}{4} \right) \, dx$

$= \left[ \frac{4x^2}{2} + 5x \right]_{-1}^{0} + \left[ 5x - \frac{x^2}{2} \right]_{0}^{3} - \frac{1}{4} \left[ \frac{x^2}{2} + 5x \right]_{-1}^{3}$

$= (0 + 3) + \left( \frac{21}{2} - 0 \right) - \frac{1}{4} \left( \frac{39}{2} + \frac{9}{2} \right)$

$= 3 + \frac{21}{2} - 6 = \frac{15}{2} \text{ units}^2$