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Question
Using the method of integration, find the area of the region bounded by the lines 3x − 2y + 1 = 0, 2x + 3y − 21 = 0 and x − 5y + 9 = 0
Solution
`3"x" - 2"y" + 1 = 0 ⇒ "y"_1 =((3"x"+1))/2` ...........(i)
`2"x" - 3"y" - 21 = 0 ⇒ "y"_2 =((21-2"x"))/3` .....(ii)
`"x" - 5"y" + 9 = 0 ⇒ "y"_3 = (("x"+9))/5` ......(iii)
Point of intersection of (i) and (ii) is A(3, 5)
Point of intersection of (ii) and (iii) is B(6, 3) and
Point of intersection of (iii) and (i) is C(1, 2).
Therefore, the area of the region bounded=`int_1^3"y"_1."dx"+int_3^6"y"_2."dx"-int_1^6"y"_3". dx"`
`=int_3^1((3"x"+1))/2."dx"+int_3^6((21-2"x"))/3."dx"-int_1^6(("x"+9))/5."dx"`
`=1/2((3"x"^2)/2+"x")_1^3+1/3(21"x"-"x"^2)_3^6-1/5("x"^2/2+9"x")_1^6`
`1/2[14]+1/3[36]-1/5[125/2]`
`= 7+12-12.5`
`= 6.5 "sq.units"`
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