## 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**

- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.4
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS
- RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS

**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:**

**(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

**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.

**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 + a**

^{2}= 0, find the values of a.**Solution:**

x = 1, y = 6 is a solution of the equation

8x – ay + a

^{2}= 0

Substituting the value of x and y,

8 x 1-a x 6 + a

^{2}= o

⇒ 8 – 6a + a

^{2}= 0

⇒ a

^{2}– 6a + 8 = 0

⇒ a

^{2}– 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**

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