=> Using the method of integration, find the area of the region bounded the lines. 3x-2y+1 = 0, 2x+ 3y - 21 =0 and x-5y +9=0.
Answers
Answer:
Given lines are
3x−2y+1=0 ...(1)
2x+3y−21=0 ...(2)
x−5y+9=0 ...(3)
For intersection of (1) and (2)
Multiplying eq.(1) by 3 and eq.(2) by 2 and adding them, we get
9x−6y+3+4x+6y−42=0
13x−39=0
x=3
Putting it in eq.(1), we get
9−2y+1=0
2y=10
y=5
Intersection point of (1) and (2) is (3,5)
For intersection of (2) and (3)
Multiplying eq.(3) by 2 and subtracting it from eq.(2), we get
2x+3y−21−2x+10y−18=0
13y−39=0
y=3
Putting y=3 in eq,(2), we get
2x+9−21=0
2x−12=0
x=6
Intersection point of (2) and (3) is (6,3).
For intersection of (1) and (3)
Multiplying (3) x 3 and subrating it from eq.(1), we get
3x−2y+1−3x+15y−27=0
13y−26=0
y=2
Putting y = 2 in (1), we get
3x−4+1=0
x=1
Intersection point of (1) and (3) is (1,2)
With the help of point of intersection we draw the graph of lines (1), (2) and (3)
shaded region is required region.
Therefore, Area of required region = ∫
1
3
2
3x+1
dx+∫
3
6
3
−2x+21
dx−∫
1
6
5
x+9
dx
=
2
3
∫
1
3
xdx+
2
1
∫
1
3
dx−
3
2
∫
3
6
dx+7∫
3
6
dx−
5
1
∫
1
6
xdx−
5
9
∫
1
6
dx
=
2
3
[
3
x
2
]
1
3
+
2
1
[x]
1
3
−
3
2
[
2
x
2
]
3
6
+7[x]
3
6
−
5
1
[
2
x
2
]
1
6
−
5
9
[x]
1
6
=
4
3
(9−1)+
2
1
(3−1)−
6
2
(36−9)+7(6−3)−
10
1
(36−1)−
5
9
(6−1)
=6+1−9+21−
2
7
−9
=10−
2
7
=
2
20−7
=
2
13
