- RD Chapter 23- The Straight Lines Ex-23.1
- RD Chapter 23- The Straight Lines Ex-23.2
- RD Chapter 23- The Straight Lines Ex-23.3
- RD Chapter 23- The Straight Lines Ex-23.4
- RD Chapter 23- The Straight Lines Ex-23.5
- RD Chapter 23- The Straight Lines Ex-23.6
- RD Chapter 23- The Straight Lines Ex-23.7
- RD Chapter 23- The Straight Lines Ex-23.8
- RD Chapter 23- The Straight Lines Ex-23.9
- RD Chapter 23- The Straight Lines Ex-23.10
- RD Chapter 23- The Straight Lines Ex-23.11
- RD Chapter 23- The Straight Lines Ex-23.12
- RD Chapter 23- The Straight Lines Ex-23.14
- RD Chapter 23- The Straight Lines Ex-23.15
- RD Chapter 23- The Straight Lines Ex-23.16
- RD Chapter 23- The Straight Lines Ex-23.17
- RD Chapter 23- The Straight Lines Ex-23.18
- RD Chapter 23- The Straight Lines Ex-23.19

RD Chapter 23- The Straight Lines Ex-23.1 |
RD Chapter 23- The Straight Lines Ex-23.2 |
RD Chapter 23- The Straight Lines Ex-23.3 |
RD Chapter 23- The Straight Lines Ex-23.4 |
RD Chapter 23- The Straight Lines Ex-23.5 |
RD Chapter 23- The Straight Lines Ex-23.6 |
RD Chapter 23- The Straight Lines Ex-23.7 |
RD Chapter 23- The Straight Lines Ex-23.8 |
RD Chapter 23- The Straight Lines Ex-23.9 |
RD Chapter 23- The Straight Lines Ex-23.10 |
RD Chapter 23- The Straight Lines Ex-23.11 |
RD Chapter 23- The Straight Lines Ex-23.12 |
RD Chapter 23- The Straight Lines Ex-23.14 |
RD Chapter 23- The Straight Lines Ex-23.15 |
RD Chapter 23- The Straight Lines Ex-23.16 |
RD Chapter 23- The Straight Lines Ex-23.17 |
RD Chapter 23- The Straight Lines Ex-23.18 |
RD Chapter 23- The Straight Lines Ex-23.19 |

Find the angles between each of the following pairs of straight lines:

(i) 3x + y + 12 = 0 and x + 2y – 1 = 0

(ii) 3x – y + 5 = 0 and x – 3y + 1 = 0

**Answer
1** :

(i) 3x + y + 12 = 0 and x+ 2y – 1 = 0

Given:

The equations of thelines are

3x + y + 12 = 0 … (1)

x + 2y − 1 =0 … (2)

Let m_{1} andm_{2} be the slopes of these lines.

m_{1} =-3, m_{2} = -1/2

Let θ be theangle between the lines.

Then, by using theformula

tan θ = [(m_{1} –m_{2}) / (1 + m_{1}m_{2})]

= [(-3 + 1/2) / (1 +3/2)]

= 1

So,

θ = π/4 or 45^{o}

∴ The acute anglebetween the lines is 45°

(ii) 3x – y + 5 = 0 and x –3y + 1 = 0

Given:

The equations of thelines are

3x − y + 5 =0 … (1)

x − 3y + 1 =0 … (2)

Let m_{1} andm_{2} be the slopes of these lines.

m_{1} =3, m_{2} = 1/3

Let θ be theangle between the lines.

Then, by using theformula

tan θ = [(m_{1} –m_{2}) / (1 + m_{1}m_{2})]

= [(3 + 1/3) / (1 +1)]

= 4/3

So,

θ = tan^{-1} (4/3)

∴ The acute anglebetween the lines is tan^{-1} (4/3).

**Answer
2** :

Given:

The equations of thelines are

2x − y + 3 =0 … (1)

x + y + 2 = 0 … (2)

Let m_{1} andm_{2} be the slopes of these lines.

m_{1} =2, m_{2} = -1

Let θ be theangle between the lines.

Then, by using theformula

tan θ = [(m_{1} –m_{2}) / (1 + m_{1}m_{2})]

= [(2 + 1) / (1 + 2)]

= 3

So,

θ = tan^{-1} (3)

∴ The acute anglebetween the lines is tan^{-1} (3).

**Answer
3** :

To prove:

The points (2, -1),(0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram

Let us assume thepoints, A (2, − 1), B (0, 2), C (2, 3) and D (4, 0) be the vertices.

Now, let us find theslopes

Slope of AB = [(2+1) /(0-2)]

= -3/2

Slope of BC = [(3-2) /(2-0)]

= ½

Slope of CD = [(0-3) /(4-2)]

= -3/2

Slope of DA = [(-1-0)/ (2-4)]

= ½

Thus, AB is parallelto CD and BC is parallel to DA.

Hence proved, thegiven points are the vertices of a parallelogram.

Now, let us find theangle between the diagonals AC and BD.

Let m_{1} andm_{2} be the slopes of AC and BD, respectively.

m_{1} =[(3+1) / (2-2)]

= ∞

m_{2} =[(0-2) / (4-0)]

= -1/2

Thus, the diagonal ACis parallel to the y-axis.

∠ODB = tan^{-1} (1/2)

In triangle MND,

∠DMN = π/2 – tan^{-1} (1/2)

∴ The angle between thediagonals is π/2 – tan^{-1} (1/2).

**Answer
4** :

Given:

Points (2, 0), (0, 3)and the line x + y = 1.

Let us assume A (2,0), B (0, 3) be the given points.

Now, let us find theslopes

Slope of AB = m_{1}

= [(3-0) / (0-2)]

= -3/2

Slope of the line x +y = 1 is -1

∴ m_{2} =-1

Let θ be theangle between the line joining the points (2, 0), (0, 3) and the line x + y =

tan θ = |[(m_{1} –m_{2}) / (1 + m_{1}m_{2})]|

= [(-3/2 + 1) / (1 +3/2)]

= 1/5

θ = tan^{-1} (1/5)

∴ The acute anglebetween the line joining the points (2, 0), (0, 3) and the line x + y = 1is tan^{-1} (1/5).

**Answer
5** :

We need to prove:

Let us assume A (x_{1},y_{1}) and B (x_{2}, y_{2}) be the given points and Obe the origin.

Slope of OA = m_{1} =y_{1×1}

Slope of OB = m_{2} =y_{2×2}

It is giventhat θ is the angle between lines OA and OB.

Hence proved.

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