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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).  

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)

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

Posted by

Safeer PP

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