Find the area of the traingle whose vertices are (-1,1),(0,5) and (3,2), using integration.

 

 

 

 
 
 
 
 

Answers (1)

Let A(-1,1), B(0,5),and c(3,2)
The equation of line AB is

y-1= \frac{5-1}{0+1}\left ( x+1 \right )
y= 4x+5 ----(1)
The equation of line BC is
y-5= \frac{2-5}{3-0}\left ( x-0 \right )
\Rightarrow 3y-15= -3x\: \: y= \frac{-3x+15}{3}\: \: \: \: y= -x+5 ---(2)
The equation of line CA is
y-2= \frac{1-2}{-1-3}\left ( x-3 \right )
y= \frac{x}{4}+\frac{5}{4}\: ---(3)
The required area 
A= \int_{-1}^{0}y_{1}dx+\int_{0}^{3}y_{2}dx-\int_{-1}^{3}y_{3}dx
solve y_{1}= 4x+5,\: \: y_{2}= -x+5,\: \: y_{3}= \frac{x+5}{4}
[from (1),(2) & (3)]
\Rightarrow A= \int_{-1}^{0}\left ( 4x+5 \right )dx+\int_{0}^{3}\left ( -x+5 \right )dx-\int_{-1}^{3}\left ( \frac{x+5}{4} \right )dx
A\Rightarrow \left [ \frac{4x^{2}}{2}+5x \right ]^{0}_{-1}+\left [ \frac{-x^{2}}{2} +5x\right ]^{3}_{0}-\frac{1}{4}\left [ \frac{x^{2}}{2} +5x\right ]^{3}_{-1}
\Rightarrow \left [ 2x^{2}+5x \right ]^{0}_{-1}+\left [ \frac{-x^{2}}{2} +5x\right ]^{3}_{0}-\frac{1}{4}\left [ \frac{x^{2}}{2}+5x \right ]^{3}_{-1}
On applying limits:-
\Rightarrow \frac{4}{2}\left [ \left ( 0-1 \right ) +5\left ( 0+1 \right )\right ]+\left [ -\frac{1}{2} \left ( 9-0 \right )+5\left ( 3-0 \right )\right ]+\frac{1}{4}\left [ \frac{1}{2} \left ( 9-1 \right )+5\left ( 3+1 \right )\right ]
\Rightarrow 3+\frac{21}{2}-6= \frac{15}{2}  sq unit is
Hence the required area= \frac{15}{2} sq units
 
 
 

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