Question

Find the area bounded by the lines y = 4x + 5, y = 5 − x and 4y = x + 5.

Answer 1

We have, 
\[y = 4x + 5 . . . . . \left( 1 \right)\]
\[y = 5 - x . . . . . \left( 2 \right)\]
\[4y = x + 5 . . . . . \left( 3 \right)\]
All the three equations represent equations of straight lines
The points of intersection is obtained by solving  simultaneous equations
\[\text{ From }\left( 1 \right)\text{ and }\left( 2 \right)\]
\[4x + 5 = 5 - x\]
\[ \Rightarrow 5x = 0\]
\[ \Rightarrow x = 0\]
\[ \Rightarrow y = 5 \]
\[\text{ Thus }A\left( 0, 5 \right)\text{ is the point of intersection of }\left( 1 \right)\text{ and }\left( 2 \right)\]
\[\text{ From }\left( 2 \right)\text{ and }\left( 3 \right)\]
\[4\left( 5 - x \right) = x + 5\]
\[ \Rightarrow 5x = 15\]
\[ \Rightarrow x = 3\]
\[ \Rightarrow y = 2\]
\[\text{ Thus }B\left( 3, 2 \right) \text{ is the point of intersection of } \left( 2 \right)\text{ and }\left( 3 \right)\]
\[\text{ From }\left( 1 \right)\text{ and }\left( 3 \right)\]
\[4\left( 4x + 5 \right) = x + 5\]
\[ \Rightarrow 15x = - 15\]
\[ \Rightarrow x = - 1\]
\[ \Rightarrow y = 1\]
\[\text{ Thus }C\left( - 1, 1 \right)\text{ is the point of intersection of }\left( 1 \right)\text{ and }\left( 3 \right)\]
\[\text{ Area }\left( ABC \right) = \text{ area }\left( ABP \right) + \text{ area }\left( PAB \right)\]
\[ = \int_{- 1}^0 \left[ \left( 4x + 5 \right) - \left( \frac{x + 5}{4} \right) \right] dx + \int_0^3 \left[ \left( 5 - x \right) - \left( \frac{x + 5}{4} \right) \right] dx\]
\[ = \int_{- 1}^0 \left( \frac{15}{4}x + \frac{15}{4} \right) dx + \int_0^3 \left( \frac{15}{4} - \frac{5}{4}x \right) dx\]
\[ = \frac{15}{4} \left[ \frac{x^2}{2} + x \right]_{- 1}^0 + \frac{5}{4} \left[ 3x - \frac{x^2}{2} \right]_0^3 \]
\[ = \frac{15}{4}\left( - \frac{1}{2} + 1 \right) + \frac{5}{4}\left( 9 - \frac{9}{2} \right)\]
\[ = \frac{15}{8} + \left( \frac{5}{4} \times \frac{9}{2} \right)\]
\[ = \frac{15}{8} + \frac{45}{8} \]
\[ = \frac{60}{8}\]
\[ = \frac{15}{2}\text{ sq . units }\]