- 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-23.13
- 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

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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 |
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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-23.13 |
RD Chapter 23- The Straight Lines Ex-23.14 |
RD Chapter 23- The Straight Lines Ex-23.15 |
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RD Chapter 23- The Straight Lines Ex-23.19 |

Find the slopes of the lines which make the following angles with the positive direction of x – axis:

(i) – π/4

(ii) 2π/3

**Answer
1** :

(i) – π/4

Let the slope of the line be ‘m’

Where, m = tan θ

So, the slope of Line is m = tan (– π/4)

= – 1

∴ The slope of the line is – 1.

(ii) 2π/3

Let the slope of the line be ‘m’

Where, m = tan θ

So, the slope of Line is m = tan (2π/3)

∴ The slope of the line is –√3

Find the slopes of a line passing through the following points :

(i) (–3, 2) and (1, 4)

(ii) (at^{2}_{1}, 2at_{1}) and (at^{2}_{2},2at_{2})

**Answer
2** :

(i) (–3, 2) and (1, 4)

By using the formula,

∴ The slope of the lineis ½.

(ii) (at^{2}_{1},2at_{1}) and (at^{2}_{2}, 2at_{2})

By using the formula,

State whether the two lines in each of the following are parallel, perpendicular or neither:

(i) Through (5, 6) and (2, 3); through (9, –2) and (6, –5)

(ii) Through (9, 5) and (– 1, 1); through (3, –5) and (8, –3)

**Answer
3** :

(i) Through (5, 6) and (2,3); through (9, –2) and (6, –5)

By using the formula,

So, m_{2} =1

Here, m_{1} = m_{2 }=1

∴ The lines areparallel to each other.

(ii) Through (9, 5) and (–1, 1); through (3, –5) and (8, –3)

By using the formula,

Find the slopes of a line

(i) which bisects the first quadrant angle

(ii) which makes an angle of 300 with the positive direction of y – axis measured anticlockwise.

**Answer
4** :

(i) Which bisects thefirst quadrant angle?

Given: Line bisectsthe first quadrant

We know that, if theline bisects in the first quadrant, then the angle must be between line and thepositive direction of x – axis.

Since, angle = 90/2 =45^{o}

By using the formula,

The slope of the line,m = tan θ

The slope of the linefor a given angle is m = tan 45^{o}

So, m = 1

∴ The slope of the lineis 1.

(ii) Which makes anangle of 30^{0} with the positive direction of y – axis measuredanticlockwise?

Given: The line makesan angle of 30^{o} with the positive direction of y – axis.

We knowthat, angle between line and positive side of axis => 90^{o} +30^{o} = 120^{o}

By using the formula,

The slope of the line,m = tan θ

The slope of the linefor a given angle is m = tan 120^{o}

So, m = – √3

∴ The slope of the lineis – √3.

Using the method of slopes show that the following points are collinear:

(i) A (4, 8), B (5, 12), C (9, 28)

(ii) A(16, – 18), B(3, – 6), C(– 10, 6)

**Answer
5** :

(i) A (4, 8), B (5, 12), C(9, 28)

By using the formula,

The slope of the line= [y_{2} – y_{1}] / [x_{2} – x_{1}]

So,

The slope of line AB =[12 – 8] / [5 – 4]

= 4 / 1

The slope of line BC =[28 – 12] / [9 – 5]

= 16 / 4

= 4

The slope of line CA =[8 – 28] / [4 – 9]

= -20 / -5

= 4

Here, AB = BC = CA

∴ The Given points arecollinear.

(ii) A(16, – 18), B(3, –6), C(– 10, 6)

By using the formula,

The slope of the line= [y_{2} – y_{1}] / [x_{2} – x_{1}]

So,

The slope of line AB =[-6 – (-18)] / [3 – 16]

= 12 / -13

The slope of line BC =[6 – (-6)] / [-10 – 3]

= 12 / -13

The slope of line CA =[6 – (-18)] / [-10 – 16]

= 12 / -13

= 4

Here, AB = BC = CA

∴ The Given points arecollinear.

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