Find the area bounded by the lines y = 4x + 5, y = 5 - x and 4y = x +…

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

Given equations of lines are

$y = 4x + 5 \quad \dots(i)$

$y = 5 - x \quad \dots(ii)$

and

$4y = x + 5 \quad \dots(iii)$

On solving Eqs. (i) and (ii), we get

$4x + 5 = 5 - x$

$\Rightarrow x = 0$

On solving Eqs. (i) and (iii), we get

$4(4x + 5) = x + 5$

$\Rightarrow 16x + 20 = x + 5$

$\Rightarrow 15x = -15$

$\Rightarrow x = -1$

On solving Eqs. (ii) and (iii), we get

$4(5 - x) = x + 5$

$\Rightarrow 20 - 4x = x + 5$

$\Rightarrow x = 3$

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

$= \left[ \frac{4x^2}{2} + 5x \right]_{-1}^{0} + \left[ 5x - \frac{x^2}{2} \right]_{0}^{3} - \frac{1}{4} \left[ \frac{x^2}{2} + 5x \right]_{-1}^{3} \quad \left[ ∵\int x^n \, dx = \frac{x^{n+1}}{n+1} \right]$

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

$= 3 + \frac{21}{2} - \frac{1}{4} \cdot 24$

$= -3 + \frac{21}{2} = \frac{15}{2} \text{ sq. units}$