Find the area of region bounded by the triangle whose vertices are (-…

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

Let we have the vertices of a $\triangle ABC$ as $A(-1, 1)$, $B(0, 5)$ and $C(3, 2)$.

First of all we find the equations of the sides of the triangle by using

$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

$∴$ Equation of $AB$ is $y - 1 = \left( \frac{5 - 1}{0 + 1} \right)(x + 1)$

$\Rightarrow y - 1 = 4x + 4$

$\Rightarrow y = 4x + 5 \text{ say } (y_1) \quad \dots(i)$

and equation of $BC$ is $y - 5 = \left( \frac{2 - 5}{3 - 0} \right)(x - 0)$

$\Rightarrow y - 5 = \frac{-3}{3}(x)$

$\Rightarrow y = 5 - x \text{ say } (y_2) \quad \dots(ii)$

Similarly, equation of $AC$ is $y - 1 = \left( \frac{2 - 1}{3 + 1} \right)(x + 1)$

$\Rightarrow y - 1 = \frac{1}{4}(x + 1)$

$\Rightarrow 4y = x + 5$

$\Rightarrow y = \frac{x + 5}{4} \text{ (Say } y_3) \quad \dots(iii)$

$∴$ Area of shaded region $= \int_{-1}^{0} (y_1 - y_3) \, dx + \int_{0}^{3} (y_2 - y_3) \, dx$

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

$= \left[ \frac{4x^2}{2} + 5x - \frac{x^2}{8} - \frac{5x}{4} \right]_{-1}^{0} + \left[ 5x - \frac{x^2}{2} - \frac{x^2}{8} - \frac{5x}{4} \right]_{0}^{3}$

$= \left[ 0 - \left( 4 \cdot \frac{1}{2} + 5(-1) - \frac{1}{8} + \frac{5}{4} \right) \right] + \left[ \left( 15 - \frac{9}{2} - \frac{9}{8} - \frac{15}{4} \right) - 0 \right]$

$= \left[ -2 + 5 + \frac{1}{8} - \frac{5}{4} + 15 - \frac{9}{2} - \frac{9}{8} - \frac{15}{4} \right]$

$= 18 + \left( \frac{1 - 10 - 36 - 9 - 30}{8} \right)$

$= 18 + \left( -\frac{84}{8} \right) = 18 - \frac{21}{2} = \frac{15}{2} \text{ sq. units}$