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प्रश्न
Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices area A(1, 2), B (2, 0) and C (4, 3).
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उत्तर
Let A(1, 2), B (2, 0) and C (4, 3) be the vertices of a triangle ABC
\[\text { Area of } \bigtriangleup ABC = \text { Area of trapezium } ADEC - \text { Area of } \bigtriangleup ADB - \text { Area of } \bigtriangleup CBE\]
\[\text { Equation of sides AC, AB and BC are given by }: \]
\[y = \frac{x}{3} + \frac{5}{3}, y = - 2x + 4 \text { and } y = \frac{3x}{2} - 3 \text {P respectively }\]
\[\text { Hence, area of } \bigtriangleup ABC = \int_1^4 \left( \frac{x}{3} + \frac{5}{3} \right)dx - \int_1^2 \left( - 2x + 4 \right)dx - \int_2^4 \left( \frac{3}{2}x - 3 \right)dx\]
\[ = \frac{1}{3} \left[ \frac{x^2}{2} + 5x \right]_1^4 - \left[ - \frac{2 x^2}{2} + 4x \right]_1^2 - \left[ \frac{3}{4} x^2 - 3x \right]_2^4 \]
\[ = \frac{1}{3}\left[ \left( \frac{16}{2} + 20 \right) - \left( \frac{1}{2} + 5 \right) \right] - \left[ \left( - 4 + 8 \right) - \left( - 1 + 4 \right) \right] - \left[ \left( 12 - 12 \right) - \left( 3 - 6 \right) \right]\]
\[ = \frac{1}{3}\left[ 28 - \frac{11}{2} \right] - \left[ 4 - 3 \right] - \left[ 0 + 3 \right]\]
\[ = \frac{1}{3}\left[ \frac{45}{2} \right] - \left[ 1 \right] - \left[ 3 \right]\]
\[ = \frac{7}{2}\]
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