Using the method of method of integration, find the area of the region bounded by the following lines:
3x – y – 3 = 0,
2x + y – 12 = 0,
x – 2y – 1 = 0

Solution

Given equations are:

3 x - y = 3 ........( i )

2 x + y = 12 ........( ii )

x - 2 y = 1 .......( iii )

To solve ( i ) and ( ii ),

( i ) + ( ii ) $\Rightarrow$ 5 x = 15 $\Rightarrow$ x = 3

( ii ) $\Rightarrow$ y = 12 - 6 = 6

Thus ( i ) and ( ii ) intersect at C ( 3, 6 ).

To solve ( ii ) and ( iii ),

( ii ) - 2 ( iii ) $\Rightarrow$ 5 y = 10 $\Rightarrow$ y = 2

( ii ) $\Rightarrow$ 2 x = 12 - 2 = 10 $\Rightarrow$ x = 5

Thus, ( ii ) and ( iii ) intersect at B ( 5, 2 ).

To solve ( iii ) and ( i ),

2 ( i ) - ( iii ) $\Rightarrow$ 5 x = 5 $\Rightarrow$ x = 1

( iii ) $\Rightarrow$ 1 - 2 y = 1 $\Rightarrow$ y = 0

Thus ( iii ) and ( i ) intersect at A ( 1, 0 ).

$Area = \int_{1}^{3} (3 x - 3) dx + \int_{3}^{5} (12 - 2 x) dx - \int_{1}^{5} \frac{1}{2} (x - 1) dx = 3 {[\frac{x^{2}}{2} - x]}_{1}^{3} + {[12 x - x^{2}]}_{3}^{5} - \frac{1}{2} {[\frac{x^{2}}{2} - x]}_{1}^{5} = 3 [(\frac{9}{2} - 3) - (\frac{1}{2} - 1)] + [(60 - 25) - (36 - 9)] - \frac{1}{2} [(\frac{25}{2} - 5) - (\frac{1}{2} - 1)]$

$= 3 [\frac{3}{2} + \frac{1}{2}] + [35 - 27] - \frac{1}{2} [\frac{15}{2} + \frac{1}{2}] = 3 [\frac{3}{2} + \frac{1}{2}] + [35 - 27] - \frac{1}{2} [\frac{15}{2} + \frac{1}{2}] = 3 [\frac{4}{2}] + [35 - 27] - \frac{1}{2} [\frac{16}{2}] = 3 [2] + [35 - 27] - \frac{1}{2} [8]$

= 6 + 8 - 4

= 10 sq. unit

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Vector Algebra

Using the method of method of integration, find the area of the region bounded by the following lines:
3x – y – 3 = 0,
2x + y – 12 = 0,
x – 2y – 1 = 0

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For Daily Free Study Material Join wiredfaculty

Vector Algebra

Using the method of method of integration, find the area of the region bounded by the following lines:3x – y – 3 = 0,2x + y – 12 = 0,x – 2y – 1 = 0

Some More Questions From Vector Algebra Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Using the method of method of integration, find the area of the region bounded by the following lines:
3x – y – 3 = 0,
2x + y – 12 = 0,
x – 2y – 1 = 0