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

This question has appeared in model paper 2012.

$\begin{array}{1 1}(A) 6\;sq.units \\ (B)3\;sq.units \\ (C) 4\;sq.units \\ (D)12 sq.units \end{array} $

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  • The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
  • Equation of the line when two points are given is $\large\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$
Step 1:
Given lines are $x+2y=2$,$y-x=1$ and $2x+y=7$.
Consider line $x+2y=2$,This meets the $x$-axis at $(2,0)$ and $y$-axis at $(0,1).$
Join these two points to obtain the graph of $x+2y=2$
Similarly the graph of other two lines can be drawn as shown in the fig.
Step 2:
Let us solve these equations to obtain the points of intersection.
$\;\;\;3y=3\Rightarrow y=1$ and $x=0$
$-3x\;\;=-6\Rightarrow x=2$ and $y=3$
$x+2y=2\;(\times\; 2)$
$\;\;\;3y=-3\Rightarrow y=-1$ and $x=4$
Hence the points of intersection are $(0,1),(2,3)$ and $(4,-1)$
The required area is $\int_0^2[(x+1)-\big(\large\frac{2-x}{2}\big)]$$dx+\int_2^4[(7-2x)-\big(\large\frac{2-x}{2}\big)]$$dx$
Step 3:
On integrating we get,
$\begin{bmatrix}x^2+x-\large\frac{1}{2}\normalsize (x-\large\frac{x^2}{2})\end{bmatrix}_0^2+\begin{bmatrix}(7x-x^2)-\large\frac{1}{2}\normalsize (x-\large\frac{x^2}{2}\big)\end{bmatrix}_2^4$
Applying the limits we get
$[4+2-\large\frac{1}{2}\normalsize (2-\frac{4}{2})-0]+\normalsize[(28-16)-\large\frac{1}{2}$$(4-\large\frac{16}{2})\normalsize -(14-4)-\frac{1}{2}\normalsize (2-\large\frac{4}{2})]$
On simplifying we get,
Hence the required area=6 sq. units.
answered May 6, 2013 by sreemathi.v
edited May 7, 2013 by sreemathi.v

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