Question

Using integration, find the area of the region bounded by the triangle ABC whose vertices A, B, C are (−1, 1), (0, 5) and (3, 2) respectively.

Answer 1

Equation of line AB is
\[y - 1 = \left( \frac{5 - 1}{0 + 1} \right)\left( x - \left( - 1 \right) \right)\]
\[ \Rightarrow y = 4x + 5 \]
Area under the line AB = area ABDO
\[ = \int_{- 1}^0 \left( 4x + 5 \right) dx\]
\[ = \left[ 4\frac{x^2}{2} + 5x \right]_{- 1}^0 \]
\[ = 0 - \left( 2 - 5 \right)\]
\[ \Rightarrow\text{ Area ABDO }= 3\text{ sq . units .} . . \left( 1 \right)\]
Equation of line BC is 
\[y - 5 = \left( \frac{2 - 5}{3 - 0} \right)\left( x - 0 \right)\]
\[ \Rightarrow y = - x + 5\]
Area under line BC = Area OBCP
\[ = \int_0^3 \left( - x + 5 \right) dx\]
\[ = \left[ - \frac{x^2}{2} + 5x \right]_0^3 \]
\[ = - \frac{9}{2} + 15 - 0\]
\[ \Rightarrow\text{ Area OBCP }= \frac{21}{2}\text{ sq . units . }. . \left( 2 \right)\]
Equation of line CA is
\[y - 2 = \left( \frac{2 - 1}{3 - \left( - 1 \right)} \right)\left( x - 3 \right) \]
\[ \Rightarrow 4y = x + 5\]
\[ \therefore\text{ Area under line AC = Area ACPA }\]
\[ \Rightarrow A = \int_{- 1}^3 \left( \frac{x + 5}{4} \right)dx\]
\[ \Rightarrow A = \frac{1}{4} \left[ \frac{x}{2}^2 + 5x \right]_{- 1}^3 \]
\[ \Rightarrow A = \frac{1}{4}\left[ \frac{3}{2}^2 + 5 \times 3 - \frac{\left( - 1 \right)}{2}^2 + 5\left( - 1 \right) \right]\]
\[ \Rightarrow A = \frac{1}{4}\left[ \frac{9}{2} + 15 - \frac{1}{2} + 5 \right]\]
\[ \Rightarrow\text{ Area ACPA }= \frac{24}{4} = 6 \text{ sq . units }. . . \left( 3 \right)\]
\[\text{ From }\left( 1 \right), \left( 2 \right)\text{ and }\left( 3 \right)\]
\[\text{ Area }\Delta\text{ ABC }=\text{ Area ABDO + Area OBCP - Area ACPA }\]
\[ \Rightarrow A = 3 + \frac{21}{2} - 6\]
\[ \Rightarrow A = \frac{21}{2} - 3 = \frac{21 - 6}{2} = \frac{15}{2} \text{ sq . units }\]
\[ \therefore\text{ Area }\Delta\text{ ABC }= \frac{15}{2}\text{ sq . units }\]