Question

Using integration, find the area of the triangle ABC coordinates of whose vertices are A (4, 1), B (6, 6) and C (8, 4).

Answer 1

A(4, 1), B(6, 6) and C(8, 4) are three given points.

Equation of AB is given by
\[y - 1 = \frac{6 - 1}{6 - 4}\left( x - 4 \right)\]
\[ \Rightarrow y - 1 = \frac{5}{2}\left( x - 4 \right)\]
\[ \Rightarrow y = \frac{5}{2}x - 9 . . . . . \left( 1 \right)\]
Equation of BC is given by
\[y - 6 = \frac{4 - 6}{8 - 6} \left( x - 6 \right)\]
\[ \Rightarrow y - 6 = - \left( x - 6 \right)\]
\[ \Rightarrow y = - x + 12 . . . . . \left( 2 \right)\]
Equation of CA is given by
\[ \Rightarrow y - 4 = \frac{1 - 4}{4 - 8}\left( x - 8 \right)\]
\[ \Rightarrow y - 4 = \frac{3}{4}\left( x - 8 \right)\]
\[ \Rightarrow y = \frac{3}{4}x - 2 . . . . . \left( 3 \right)\]
Required area of shaded region ABC
\[\text{ Shaded area }\left( ABC \right) = \text{ area }\left( ABD \right) \hspace{0.167em} + area\left( DBC \right)\]
\[\text{ Consider a point }P\left( x, y_2 \right)\text{ on AB and }Q\left( x, y_1 \right)\text{ on AD }\]
\[ \Rightarrow\text{ for a vertical strip of length }= \left| y_2 - y_1 \right|\text{ and width }= dx, \text{ area }= \left| y_2 - y_1 \right|dx, \]
\[\text{ The approximating rectangle moves from } x = 4\text{ to }x = 6\]
\[\text{ Hence area }\left( ABD \right) = \int_4^6 \left| y_2 - y_1 \right|dx\]
\[ = \int_4^6 \left[ \left( \frac{5}{2}x - 9 \right) - \left( \frac{3}{4}x - 2 \right) \right]dx\]
\[ = \int_4^6 \left( \frac{7}{4}x - 7 \right)dx\]
\[ = \left[ \frac{7}{4} \times \frac{x^2}{2} - 7x \right]_4^6 \]
\[ = \frac{7}{8}\left( 36 - 16 \right) - 7\left( 6 - 4 \right)\]
\[ = \frac{7}{8} \times 20 - 14\]
\[ = \frac{35}{2} - 14\]
\[ = \frac{35 - 28}{2} = \frac{7}{2}\text{ sq units }\]
\[\text{ Consider a point S }\left( x, y_4 \right)\text{ on BC and R }\left( x, y_3 \right)\text{ on DC }\]
\[\text{ For a vertical strip of length }= \left| y_4 - y_3 \right|\text{ and width }= dx, \text{ area }= \left| y_4 - y_3 \right|dx, \]
\[\text{ The approximating rectangle moves from }x = 6\text{ to } x = 8\]
\[ \Rightarrow \text{ area }\left( BDC \right) = \int_6^8 \left[ \left( - x + 12 \right) - \left( \frac{3}{4}x - 2 \right) \right] dx\]
\[ = \int_6^8 \left( - \frac{7}{4}x + 14 \right)dx\]
\[ = \left[ - \frac{7}{4} \times \frac{x^2}{2} + 14x \right]_6^8 \]
\[ = \left[ - \frac{7 x^2}{8} + 14x \right]_6^8 \]
\[ = - \frac{7}{8}\left( 64 - 36 \right) + 14\left( 8 - 6 \right)\]
\[ = - \frac{7}{8} \times 28 + 28\]
\[ = \frac{28}{8} = \frac{7}{2}\]
\[\text{ Thus required area }\left( ABC \right) = \frac{7}{2} + \frac{7}{2} = 7 \text{ sq . units }\]