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For the given equations, determine graphically the vertices of the triangle, (i)2b-a=8, 5b-a =14 and b-2a=1 (ii) b=a, b=0 and 3a+3b=10
CBSE | Class 10 | Exercise 3.2 | Maths | Pair of Linear Equations In Two Variables | RD Sharma
For the given equations, determine graphically the vertices of the triangle, (i)2b-a=8, 5b-a =14 and b-2a=1 (ii) b=a, b=0 and 3a+3b=10

Given,

2b-a=8……. (i)

5b-a=14……. (ii)

b-2a=1……… (iii)

For equation (i),

⇒ b=(a+8)/2

When a=-4, we have b=(-4+8)/2=2

When a=0, we have b=(0+8)/2=4

a -4 0
b 2 4

Thus, we have the following table giving points on the line 2b-a=8

From equation (ii),

Solve for b:

⇒ b=(a+14)/5

So, when a=-4

b=((-4)+14)/5=2

And, when a=1

⇒ b=(1+14)/5=3

Thus, we have the following table giving points on the line 5b-a=14

a -4 1
b 2 3

Finally, for equation (iii),

⇒ b=(2a+1)

When a=-1, we have b=(2(-1)+1)=-1

When a=1, we have b=(2(1)+1)=3

Thus, we have the following table giving points on the line b-2a=1

a -1 1
b 1 3

Graph of the equations (i), (ii) and (iii) is given below:

Given,

b=a ……. (i)

b=0 ……. (ii)

3a+3b=10……… (iii)

From equation (i),

When a=1, we have b=1

When a=-2, we have b=-2

Thus, we have the following table giving points on the line b=a

a 1 -2
b 1 -2

From equation (ii),

When a=0

b=0

And, when a=10/3

⇒ b=0

Thus, we have the following table giving points on the line b=0

a 0 10/3
b 0 10/3

Finally, for equation (iii),

⇒ b=(10-3a)/3

When a=1, we get b=(10-3(1))/3)=7/3

When a=2, we get b=(10-3(2))/3=4/3

a 1 2
b 7/3 4/3

Thus, we have the following table giving points on the line 3a+3b=10

Graph of the equations (i), (ii) and (iii) is given below:

From the above graph, we observe that the lines taken in pairs intersect at points A(0,0) B(10/3,0) and C(5/3,5/3)

Hence the vertices of the triangle are A(0,0) B(10/3,0) and C(5/3,5/3).

i

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