RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2

Other Exercises

Question 1.
Write two solutions for each of the following equations
(i) 3x + 4y = 7
(ii) x = 6y
(iii) x + πy = 4
(iv) \(\frac { 2 }{ 3 }\) x – y = 4
Solution:

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2 Q1.1
(ii) x = 6y
Let y = 0, then
x = 6 x 0 = 0
∴ x = 0, y = 0
x = 0, y = 0 are the solutions of the equation
Let y= 1, then
x = 6 x 1 = 0 –
∴ x = 6, y = 1 are the solutions of the equation.
(iii) x + πy = 4
Let x = 4, then
4 + πy = 4
⇒ πy = 4- 4 = 0
∴ y = 0
∴ x = 4, y = 0 are the solutions of the equation
Let x = 0, then
0 + πy = 4 ⇒ πy = 4
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2 Q1.2

Question 2.
Check which of the following are solutions of the equations 2x – y =6 and which are not
(i) (3, 0)
(ii) (0, 6)
(iii) (2,-2)
(iv)(\(\sqrt { 3 } \) ,0)
(v) (\(\frac { 1 }{ 2 }\) ,-5)
Solution:
Equation is 2x – y = 6
(i) Solution is (3, 0) i.e. x = 3, y = 0
Substituting the value of x and y in the equation
2 x 3 – 0 = 6 ⇒ 6 – 0 = 6
6 = 6
Which is true
∴ (3, 0) is the solutions.
(ii) (0, 6) i.e. x =0, y =6
Substituting the value of x and y in the equation
2 x 0 – 6 = 6 ⇒ 0-6 = 6
⇒ -6 = 6 which is not true
∴ (0, 6) is not its solution’
(iii) (2, -2) i.e. x = 2, y = -2
Substituting the value of x and y in the equation
2 x 2 – (-2) = 6 ⇒ 4 + 2 = 6
⇒ 6 = 6 which is true.
∴ (2, -2) is the solution.
(iv) (\(\sqrt { 3 } \),0) i.e. x = \(\sqrt { 3 } \) , y = 0,
Substituting the value of x and y in the equations
2 x \(\sqrt { 3 } \)-(0) = 6
⇒ 2\(\sqrt { 3 } \)-0 = 6
⇒ 2 \(\sqrt { 3 } \) =  6 which is not true
∴ (\(\sqrt { 3 } \) > 0) is not the solution.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2 Q2.1

Question 3.
If x = -1, y = 2 is a solution of the equation 3x + 4y =k Find the value of k.
Solution:
x = -1, y = 2
The equation is 3x + 4y = k
Substituting the value of x and y in it
3 x (-1) + 4 (2) = k
⇒ -3+ 8 = k
⇒ 5 = k
∴ k = 5

Question 4.
Find the value of λ if x = -λ and y = \(\frac { 5 }{ 2 }\) is a solution of the equation x + 4y – 7 = 0.
Solution:
x = -λ, y= \(\frac { 5 }{ 2 }\)
Equation is x + 4y – 7 = 0
Substituting the value of x and y,
-λ + 4 x \(\frac { 5 }{ 2 }\) -7 = 0
⇒ -λ + 10 – 7 = 0
⇒ -λ +3 = 0
∴ -λ = -3
⇒ λ = 3
Hence λ = 3

Question 5.
If x = 2α + 1 and y = α – 1 is a solution of the equation 2x – 3y + 5 = 0, find the value of α.
Solution:
x = 2α + 1, y = α – 1
are the solution of the equation 2x – 3y + 5 – 0
Substituting the value of x and y
2(2α + 1) -3 (α – 1) + 5 = 0
⇒ 4α+ 2-3α+ 3 + 5 = 0
⇒ α+10 = 0
⇒ α = -10
Hence α = -10

Question 6.
If x = 1, and y = 6 is a solution of the equation 8x – ay + a² = 0, find the values of a.
Solution:
x = 1, y = 6 is a solution of the equation
8x – ay + a² = 0
Substituting the value of x and y,
8 x 1-a x 6 + a² = o
⇒ 8 – 6a + a² = 0
⇒ a² – 6a + 8 = 0
⇒ a² – 2a -4a + 8 = 0
⇒ a (a – 2) – 4 (a – 2) = 0
⇒ (a – 2) (a – 4) = 0
Either a – 2 = 0, then a = 2
or a – 4 = 0, then a = 4
Hence a = 2, 4

Question 7.
Write two solutions of the form x = 0, y = a and x = b, y = 0 for each of the following equations.
(i) 5x – 2y = 10
(ii) -4x + 3y = 12
(iii) 2x + 3y = 24
Solution:
(i) 5x – 2y = 10
Let x = 0, then
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2 Q7.1