Compute the area bounded by the lines $x + 2y = 2$, $y - x = 1$ and $2x + y = 7$.

$6$ square units

The correct answer is Option (2) → $6$ square units

We have,

$x + 2y = 2 \quad \dots(i)$

$y - x = 1 \quad \dots(ii)$

and $2x + y = 7 \quad \dots(iii)$

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

$y - (2 - 2y) = 1 \Rightarrow 3y - 2 = 1 \Rightarrow y = 1$

On solving Eqs. (ii) and (iii), we get $2(y - 1) + y = 7$

$\Rightarrow 2y - 2 + y = 7 \Rightarrow y = 3$

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

$2(2 - 2y) + y = 7$

$\Rightarrow 4 - 4y + y = 7 \Rightarrow -3y = 3 \Rightarrow y = -1$

$∴$ Required area $= -\int\limits_{-1}^{1} (2 - 2y) \, dy + \int\limits_{-1}^{3} \frac{(7 - y)}{2} \, dy - \int\limits_{1}^{3} (y - 1) \, dy$

$= -\left[ 2y - \frac{2y^2}{2} \right]_{-1}^{1} + \left[ \frac{7y}{2} - \frac{y^2}{2 \cdot 2} \right]_{-1}^{3} - \left[ \frac{y^2}{2} - y \right]_{1}^{3}$

$= -\left[ 2 - 1 - (-2 - 1) \right] + \left[ \left( \frac{21}{2} - \frac{9}{4} \right) - \left( -\frac{7}{2} - \frac{1}{4} \right) \right] - \left[ \left( \frac{9}{2} - 3 \right) - \left( \frac{1}{2} - 1 \right) \right]$

$= -[4] + \left[ \frac{42 - 9 + 14 + 1}{4} \right] - \left[ \frac{9 - 6 - 1 + 2}{2} \right]$

$= -4 + 12 - 2 = 6 \text{ sq. units}$