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Q : 13 Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are9\small A(2,0),B(4,5)\hspace{1mm}and\hspace{1mm}C(6,3).

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Equation of line joining A and B is

\\\frac{y-0}{x-2}=\frac{5-0}{4-2}\\ y=\frac{5x}{2}-5

Equation of line joining B and C is

\\\frac{y-5}{x-4}=\frac{5-3}{4-6}\\ y=9-x

Equation of line joining A and C is

\\\frac{y-0}{x-2}=\frac{3-0}{6-2}\\ y=\frac{3x}{4}-\frac{3}{2}

ar(ABC)=ar(ABL)+ar(LBCM)-ar(ACM)

\\ar(ABL)=\int_{2}^{4}(\frac{5x}{2}-5)dx\\ =[\frac{5x^{2}}{4}-5x]_{2}^{4}\\ =(20-20)-(5-10)\\ =5\ units

\\ar(LBCM)=\int_{4}^{6}(9-x)dx\\ =[9x-\frac{x^{2}}{2}]_{4}^{6}\\ =(54-18)-(36-8)\\ =8\ units

\\ar(ACM)=\int_{2}^{6}(\frac{3x}{4}-\frac{3}{2})dx\\ =[\frac{3x^{2}}{8}-\frac{3x}{2}]_{2}^{6}\\ =(\frac{27}{2}-9)-(\frac{3}{2}-3)\\ =6\ units

ar(ABC)=8+5-6=7

Therefore the area of the triangle ABC is 7 units.

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Sayak

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