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Question

Using the method of integraton find the area of the region bounded by lines: $$2x+y=4, 3x-2y=6$$ and $$x-3y+5=0$$


Solution

The given equations of lines are
$$2x+y=4 ......... (1)$$
$$3x-2y=6 ......... (2)$$
And, $$x-3y+5=0 .......... (3)$$
The area of the region bounded by the lines is the area of $$\Delta ABC$$. $$AL$$ and $$CM$$ are the perpendicular on $$x$$-axis.
$$Area (\Delta ABC) = Area (ALMCA) - Area (ALB) - Area (CMB)$$
$$=\displaystyle \int_{1}^{4}\left (\dfrac {x+5}{3}\right )dx-\int_{1}^{2}(4-2x)dx-\int_{2}^{4}\left (\dfrac {3x-6}{2}\right )dx$$
$$=\dfrac {1}{3}\left [\dfrac {x^2}{2}+5x\right ]_{1}^{4}-[4x-x^2]_{1}^{2}-\dfrac {1}{2}\left [\dfrac {3x^2}{2}-6x\right ]_{2}^{4}$$
$$=\dfrac {1}{3}\left [8+20-\dfrac {1}{2}-5\right ]-[8-4-4+1]-\dfrac {1}{2}[24-24-6+12]$$
$$=\left (\dfrac {1}{3}\times \dfrac {45}{2}\right )-(1)-\dfrac {1}{2}(6)$$
$$=\dfrac {15}{2}-1-3$$
$$=\dfrac {15}{2}-4=\dfrac {15-8}{2}=\dfrac {7}{2}$$ sq. units

398397_428556_ans_f57b59241e60417c80395aa4aeeeea0d.png

Mathematics
RS Agarwal
Standard XII

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