Using integration, find the area of triangle ABC bounded by the lines 4x-y+5= 0,x+y-5= 0\; and\; x-4y+5= 0\cdot

 

 

 

 
 
 
 
 

Answers (1)

Let AB: 4x+y+5= 0                                  
       BC: x+y-5= 0\: and
       CA: x-4y+5= 0
On solving we get A(-1,1) B(0,5) and C(3,2)  

\therefore as\: \left ( ABC \right )= \int_{-1}^{0}\left ( 4x+5 \right )dx+\int_{0}^{3}\left ( 5-x \right )dx\; \frac{-1}{4}\int_{-1}^{3}\left ( x+5 \right )dx
= \frac{1}{8}\left [ \left ( 4x+5 \right )^{2} \right ]^{0}_{-1}-\frac{1}{2}\left [ \left ( 5-x \right )^{2} \right ]^{3}_{0}-\frac{1}{8}\left [ \left ( x+5 \right )^{2} \right ]^{3}_{-1}
= \frac{1}{8}\left [ 25-1 \right ]-\frac{1}{2}\left [ 4-25 \right ]-\frac{1}{8}\left [ 64-16 \right ]
= 3+\frac{21}{2}-6\; \; = \frac{15}{2}sq\cdot units

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