RD Chapter 7- Introduction to Euclid-s Geometry Ex-7.1 |
RD Chapter 7- Introduction to Euclid-s Geometry Ex-7.3 |
RD Chapter 7- Introduction to Euclid-s Geometry Ex-7.4 |
RD Chapter 7- Introduction to Euclid-s Geometry Ex-VSAQS |
RD Chapter 7- Introduction to Euclid-s Geometry Ex-MCQS |

Write two solutions for each of the following equations

(i) 3x + 4y = 7

(ii) x = 6y

(iii) x + πy = 4

**Answer
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

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)( ,0)

(v) (1/2 ,-5)

**Answer
2** :

Equationis 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 = 6which is true.

∴ (2, -2) is the solution.**(iv)** (,0) i.e. x = , y = 0,

Substituting the value of x and y in the equations

**Answer
3** :

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

**Answer
4** :

x= -λ, y=

Equation is x + 4y – 7 = 0

Substituting the value of x and y,

-λ + 4 x -7 = 0

⇒ -λ + 10 – 7 = 0

⇒ -λ +3 = 0

∴ -λ = -3

⇒ λ = 3

Hence λ = 3

**Answer
5** :

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

**Answer
6** :

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

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

**Answer
7** :

(i) 5x – 2y = 10

Let x = 0, then

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