Q. Compute the area bounded by the lines x + 2y =2, y - x =1 and 2x + y = 7.
full explanation..
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Answers
Answer:
Given lines are x+2y=2 ...(1)
y+2x=7 ...(2)
y−x=1 ...(3)
I) x+2y=2
x 0 2
y 1 0
(0,1);(2,0)
II) 2x+y=7
x 0
2
7
y 7 0
(0,7);(
2
7
,0)
III) y−x=1
x 0 −1
y 1 0
(0,1);(−1,0)
Now, we find points of intersection
1) (I) and (II)
x+2y=2,y+2x=7
y=7−2x
⇒x+2[7−2x]=2
x+14−4x=2
−3x=2−14=−12
x=
3
12
=4
y=7−2(4)=7−8=−1
∴ point A(4,−1)
2) (II) and (III)
y+2x=7, y−x=1
y=x+1
⇒x+1+2x=7
3x=6
x=2
⇒y=2+1=3
Point B(2,3)
3) (I) and (III)
x+2y=2, y−x=1
y=x+1
⇒x+2[x+1]=2
x+2x+2=2
3x=0
x=0
⇒y=0+1=1
Point C(0,1)
We draw graph [Ref. image]
To find shaded tregion
ar(△ABC)
=∫
0
2
y
CB
dx+∫
4
2
y
BA
.dx−∫
4
0
y
AC
.dx
=∫
0
2
(x+1)dx+∫
0
2
(7−2x)dx−∫
4
0
(
2
2−x
)dx
=[
2
x
2
+x]
0
2
+[7x−
2
2x
2
]
4
2
−
2
1
[2x−
2
x
2
]
4
0
=[
2
2
2
+2]−[
2
o
2
+0]+[7(2)−
2
2(2)
2
]−[7(4)−
2
2(4)
2
]−
2
1
[2(0)−
2
0
2
]+
2
1
[2(4)−
2
4
2
]
=[2+2]−[0]+[14−4]−[28−16]−
2
1
[0]+
2
1
[8−8]
=4+10−12+0
=14−12
=2sq.units
Step-by-step explanation:
hope it helps you
Step-by-step explanation:
Given lines are x+2y=2 ...(1)
y+2x=7 ...(2)
y−x=1 ...(3)
I) x+2y=2
x 0 2
y 1 0
(0,1);(2,0)