Question

Using the 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.

Answer 1

We have,
\[3x - y - 3 = 0 \cdots\left( 1 \right)\]
\[2x + y - 12 = 0 . . . \left( 2 \right)\]
\[x - 2y - 1 = 0 . . . \left( 3 \right)\]
\[\text{ Solving }\left( 1 \right)\text{ and }\left( 2 \right),\text{ we get, }\]
\[5x - 15 = 0 \]
\[ \Rightarrow x = 3 \]
\[ \therefore y = 6\]
\[B(3, 6)\text{ is point of intersection of }\left( 1 \right)\text{ and } \left( 2 \right)\]
\[\text{ Solving }\left( 1 \right)\text{ and }\left( 3 \right),\text{ we get, }\]
\[5x = 5\]
\[ \Rightarrow x = 1 \]
\[ \therefore y = 0\]
\[A\left( 1, 0 \right)\text{ is point of intersection of }\left( 1 \right)\text{ and }\left( 3 \right)\]
\[\text{ Solving }\left( 2 \right)\text{ and }\left( 3 \right),\text{ we get, }\]
\[5x = 25 \]
\[ \Rightarrow x = 5 \]
\[ \therefore y = 2\]
\[C\left( 5, 2 \right)\text{ is point of intersection of }\left( 2 \right)\text{ and }\left( 3 \right)\]
Now, 
\[\text{ Area ABC }= \left\{\text{ area bound by }\left( 1 \right) \text{ between }x = 1\text{ and }x = 3 \right\} + \left\{\text{ area bound by }\left( 2 \right) between x = 3\text{ and }x = 5 \right\} - \left\{\text{ area bound by }\left( 3 \right) \text{ between }x = 1\text{ and }x = 5 \right\} \]
\[ = \int_1^3 \left( 3x - 3 \right) dx + \int_3^5 \left( 12 - 2x \right)dx - \int_1^5 \frac{\left( x - 1 \right)}{2}dx\]
\[ = \left[ 3 \times \frac{x^2}{2} - 3x \right]_1^3 + \left[ 12x - 2 \times \frac{x^2}{2} \right]_3^5 - \frac{1}{2} \left[ \frac{x^2}{2} - x \right]_1^5 \]
\[ = 3 \left[ \frac{x^2}{2} - x \right]_1^3 + \left[ 12x - x^2 \right]_3^5 - \frac{1}{2} \left[ \frac{x^2}{2} - x \right]_1^5 \]
\[ = 3\left( \frac{9}{2} - 3 \right) - 3\left( \frac{1}{2} - 1 \right) + \left( 12 - 25 \right) - \left( 12 - 9 \right) - \frac{1}{2}\left[ \left( \frac{25}{2} - 1 \right) - \left( \frac{1}{2} - 1 \right) \right]\]
\[ = 6 + 8 - 4\]
\[ = 10 \text{ sq units }\]