Using method of integration, find the area of the triangle whose vertices are (1,0),(2,2)\: \: and\: \: (3,1)

 

 

 

 
 
 
 
 

Answers (1)

Let A(1,0),B(2,2)\: \: and\:\:C(3,1) be the vertices of a triangle ABC


Area\:\:of \:\:\bigtriangleup ABc=Area\:\:of\:\:trepezium \:BDEC-Area\:\:of\:: \bigtriangleup AEC
Now, the equation of side AB,

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1) \\ y-0=2(x-1)\\ y=2(x-1) \:\: -(i) \\ Equation \:\: of \:\: line \:\: BC \\ y-2= \frac{1-2}{3-2}(x-2) \\ y-2= \frac{-1}{1}(x-2)\\ y-2 = - (x-2) \Rightarrow y=2-(x-2)\\ y=4-x \:\:\: -(ii) \\ Equation \:\: of \:\:line \:\: AC \\ y-0=\frac{1-0}{3-1}(x-1)\\ = \frac{1}{2}(x-1) \\ y=\frac{1}{2}(x-1) \:\:\: - (iii)

Hence the area of  \bigtriangleup ABC \\

=\int_{1}^{2}2(x-1)dx+\int_{2}^{3}(4-x)dx-\int_{1}^{3}\frac{x-1}{2}dx \\ = 2 \left [ \frac{x^2}{2}-x \right ]_{1}^{2}+ \left [ 4x-\frac{x^2}{2} \right ]_{2}^{3}- \frac{1}{2}\left [ \frac{x^2}{2}-x \right ]_{1}^{3} \\ = 2 \left [ \left (\frac{2^2}{2}-2 \right )-\left ( \frac{1}{2}-1 \right )+\left ( 4 \times 3 - \frac{3^2}{2} \right )-\left ( 4\times 2 - \frac{2^2}{2} \right )\right ]-\frac{1}{2}\left [ \left ( \frac{3^2}{2} -3 \right )-\left ( \frac{1}{2} \right )-1 \right ] \\ = 2\left ( \frac{1}{2} \right )+\frac{3}{2}-1 \: \:= \frac{3}{2}sq.units

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