- 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-23.8
- 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

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-23.8 |
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 |

Find the equation of a line for which

(i) p = 5, α = 60°

(ii) p = 4, α = 150°

**Answer
1** :

(i) p = 5, α = 60°

Given:

p = 5, α = 60°

The equation of theline in normal form is given by

Using the formula,

x cos α + y sin α= p

Now, substitute thevalues, we get

x cos 60° + y sin 60°= 5

x/2 + √3y/2 = 5

x + √3y = 10

∴ The equation of linein normal form is x + √3y = 10.

(ii) p = 4, α = 150°

Given:

p = 4, α = 150°

The equation of theline in normal form is given by

Using the formula,

x cos α + y sin α= p

Now, substitute thevalues, we get

x cos 150° + y sin150° = 4

cos (180° – θ) =– cos θ , sin (180° – θ) = sin θ

x cos(180° – 30°)+ y sin(180° – 30°) = 4

– x cos 30° + ysin 30° = 4

–√3x/2 + y/2 = 4

-√3x + y = 8

∴ The equation of linein normal form is -√3x + y= 8.

**Answer
2** :

Given:

p = 4, α = 30°

The equation of theline in normal form is given by

Using the formula,

x cos α + y sin α= p

Now, substitute thevalues, we get

x cos 30° + y sin 30°= 4

x√3/2 + y1/2 = 4

√3x + y = 8

∴ The equation of linein normal form is √3x + y = 8.

**Answer
3** :

Given:

p = 4, α = 15°

The equation of theline in normal form is given by

We know that, cos15° = cos (45° – 30°) = cos45°cos30° + sin45°sin30°

Cos (A – B) = cos Acos B + sin A sin B

So,

And sin 15 = sin(45° – 30°) = sin 45° cos 30° – cos 45° sin 30°

Sin (A – B) = sin Acos B – cos A sin B

So,

Now, by using theformula,

x cos α + y sin α= p

Now, substitute thevalues, we get

(√3+1)x +(√3-1) y =8√2

∴ The equation of linein normal form is (√3+1)x +(√3-1) y = 8√2.

**Answer
4** :

Given:

p = 3, α = tan^{-1} (5/12)

So, tan α = 5/12

sin α = 5/13

cos α = 12/13

The equation of theline in normal form is given by

By using the formula,

x cos α + y sin α= p

Now, substitute thevalues, we get

12x/13 + 5y/13 = 3

12x + 5y = 39

∴ The equation of linein normal form is 12x + 5y = 39.

**Answer
5** :

Given:

p = 2, sin α = 1/3

We know that cos α =√(1 – sin^{2} α)

= √(1 – 1/9)

= 2√2/3

The equation of theline in normal form is given by

By using the formula,

x cos α + y sin α= p

Now, substitute thevalues, we get

x2√2/3 + y/3 = 2

2√2x + y = 6

∴ The equation of linein normal form is 2√2x + y = 6.

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