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  • Using the method od integration, find the area of the triangular region whose vertices are (2, -2), (4,3) and (1,2)        </di

Using the method od integration, find the area of the triangular region whose vertices are (2, -2), (4,3) and (1,2)

 

 

 

 
 
 
 
 

Answers (1)

Let A(2,-2), B(4,3) and C(1,2).

Equations of side AB, BC and CA are resp. 

x=\frac{2y+14}{5},x=3y-5,x=\frac{6-y}{4}

So, ar(ABC)=\int_{-2}^{3}x_{AB}dy-\int_{-2}^{2}x_{CA}dy-\int_{2}^{3}x_{BC}dy

\Rightarrow \int_{-2}^{3}\left ( \frac{2y+14}{5} \right )dy-\int_{-2}^{5}\left ( \frac{6-y}{4} \right )dy-\int_{2}^{3}(3y-5)\times 1dy

\Rightarrow \int_{-2}^{3}\left ( \frac{2y+14}{5} \right )dy-\int_{-2}^{5}\left ( \frac{6-y}{4} \right )dy-\int_{2}^{3}(3y-5)\times 1dy

\Rightarrow \frac{1}{5}\left [ (y+7)^2 \right ]_2^3+\frac{1}{8}\left [ (6-y)^2 \right ]_{-2}^2-\frac{1}{6}\left [ \left ( 3y-5 \right )^2 \right ]_2^1

\Rightarrow \frac{1}{5}\left [ 100-25 \right ]+\frac{1}{8}\left [16-64 \right ]-\frac{1}{6}\left [ 16-1 \right ]

\Rightarrow 15-6-\frac{15}{2}=\frac{13}{2} \; sq. \; units

 

Posted by

Ravindra Pindel

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