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  2. Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B(4, 5) and C(6, 3).  
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Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B(4, 5) and C(6, 3).

 

Answers (1)
S safeer

Given:

A(2, 0), B(4, 5), C(6, 3)

Required area = Area of triangle ABC

\\$ Equation of AB:$ \\\\ {y}= \frac{5-0}{4-2}(x-2) \\\\ y = \frac{5}{2}(x-2)

\\$ Equation of BC:$ \\\\ {y-5}= \frac{3-5}{6-4}(x-4) \\\\ y = 9-x

\\$ Equation of AC:$ \\\\ {y}= \frac{3-0}{6-2}(x-2) \\\\ y = \frac{3}{4}(x-2)

\\ \text {Area of trianlge ABC} =\frac{5}{2} \int_{2}^{4}(\mathrm{x}-2) \mathrm{d} \mathrm{x}+\int_{4}^{6}(9-\mathrm{x}) \mathrm{dx}-\frac{3}{4} \int_{2}^{6}(\mathrm{x}-2) \mathrm{dx} \\ \\ =\left| \frac{5}{4} (\mathrm{x}-2)^{2}\right|_{2} ^{4}+\left|\frac{-1}{2}(9-\mathrm{x})^{2}\right|_{4} ^{6}-\left|\frac{3}{8} (\mathrm{x}-2)^{2} \right|_{2} ^{6} \\\\ = \frac{5}{4}( 4-0)+\frac{-1}{2}(9-25)-\frac{3}{8} (16-0) \\\\ = 5 +8 -6= 7\ \ sq. unit

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