Answer :

To find region enclosed by

3x – y – 3 = 0 ...(i)


2x + y – 12 = 0 ...(ii)


x – 2y – 1 = 0 ...(iii)


Solving (i) and (ii), we get,


5x – 15 = 0


Or x = 3


y = 6


The points of intersection of (i) and (ii) is B (3,6)


Solving (i) and (iii), we get,


5x = 5


Or x = 1


y = 0


The points of intersection of (i) and (iii) is A (1,0)


Solving (ii) and (iii), we get,


5x = 25


Or x = 5


y = 2


The points of intersection of (ii) and (iii) is C (5,2) .


These are shown in the graph below: -



Area of the bounded region


=




= 11 sq. units


The area of the region bounded by the following line 3x – y 3 = 0, 2x + y – 12 = 0, x – 2y – 1 = 0 is


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