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

$(A)\; 5 sq.units$
$(B)\; 12 sq.units$
$(C)\; 10 sq.units$
$(D)\; 6 sq.units$

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A)
Given
$x +2y = 2, y = x +1 ,2x +y= 7$
solving equation (1) and (2)
$x +2y = 2 \;and \;y = x +1$
$x +2 (x+1) = 2 \implies x = 0 \;and \;y = 1$
solving equation (2) and (3)
$2x + (x+1) = 7 \implies x = 2, y = 3$
solving equation (3) and (1)
$(2) x +2y = 2$
$\implies y = -1$
and $x = 4$
The required area = area ABC + area ADC
=\begin{align*}\int _0^2 \begin{bmatrix} (1+x ) -( \frac{2-x}{2 } )\end{bmatrix} dx + \int_2^4 \begin{bmatrix} (7-2x) - (\frac{2-x}{2}) \end{bmatrix} dx \end{align*}
= \begin{align*} \int_0^2 \frac{3}{2} x dx + \int _2^4 (6 - \frac{3}{2}x) dx \end{align*}
$= \begin{bmatrix}\frac{3x^2}{2} \end{bmatrix} _0^2 + \begin{bmatrix} 6x - \frac{3x^2}{2} \end{bmatrix}_2^4$
$= \frac{3}{4} \times 4 + \begin{bmatrix} (24 - \frac{3 \times 16}{2}) - (6 \times 2 - \frac{3 \times 2 \times 2}{4}) \end{bmatrix}$
$= 3 + [12-9] = 6 sq.units$