- 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-23.9
- 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-23.7 |
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 |

**Answer
1** :

Given:

(x_{1}, y_{1})= A (1, 2), θ = 60°

Let us find thedistance AP.

By using the formula,

The equation of theline is given by:

Here, r represents thedistance of any point on the line from point A (1, 2).

The coordinate of anypoint P on this line are (1 + r/2, 2 + √3r/2)

It is clear that, Plies on the line x + y = 6

So,

∴ The value of AP is3(√3 – 1)

**Answer
2** :

Given:

(x_{1}, y_{1})= A (3, 4), θ = π/6 = 30°

Let us find the lengthPQ.

By using the formula,

The equation of theline is given by:

x – √3 y + 4√3 – 3 = 0

Let PQ = r

Then, the coordinateof Q are given by

A straight line drawn through the point A (2, 1) making anangle π/4 with positive x–axis intersects another line x + 2y + 1 = 0in the point B. Find length AB.

**Answer
3** :

Given:

(x_{1}, y_{1})= A (2, 1), θ = π/4 = 45°

Let us find the lengthAB.

By using the formula,

The equation of theline is given by:

x – y – 1 = 0

Let AB = r

Then, the coordinateof B is given by

A line a drawn through A (4, – 1) parallel to the line 3x – 4y + 1 = 0.Find the coordinates of the two points on this line which are at a distance of5 units from A.

**Answer
4** :

Given:

(x_{1}, y_{1})= A (4, -1)

Let us findCoordinates of the two points on this line which are at a distance of 5 unitsfrom A.

Given: Line 3x – 4y +1 = 0

4y = 3x + 1

y = 3x/4 + 1/4

Slope tan θ = 3/4

So,

Sin θ = 3/5

Cos θ = 4/5

The equation of theline passing through A (4, −1) and having slope ¾ is

By using the formula,

The equation of theline is given by:

3x – 4y = 16

Here, AP = r = ± 5

Thus, the coordinatesof P are given by

x

= ±4 + 4 and y = ±3 –1

x = 8, 0 and y = 2, –4

∴ The coordinates ofthe two points at a distance of 5 units from A are (8, 2) and (0, −4).

**Answer
5** :

Given:

The equation of theline that passes through P(x_{1}, y_{1}) and makes anangle of θ with the x–axis.

Let us find the lengthof PQ.

By using the formula,

The equation of theline is given by:

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