- 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-23.13
- 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-23.17
- 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-23.10 |
RD Chapter 23- The Straight Lines Ex-23.11 |
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RD Chapter 23- The Straight Lines Ex-23.19 |

Determine the distance between the following pair of parallel lines:

(i) 4x – 3y – 9 = 0 and 4x – 3y – 24 = 0

(ii) 8x + 15y – 34 = 0 and 8x + 15y + 31 = 0

**Answer
1** :

(i) 4x – 3y – 9 = 0 and 4x– 3y – 24 = 0

Given:

The parallel lines are

4x − 3y − 9= 0 … (1)

4x − 3y − 24= 0 … (2)

Let d be the distancebetween the given lines.

So,

∴ The distance betweengivens parallel line is 3units.

(ii) 8x + 15y – 34 = 0 and8x + 15y + 31 = 0

Given:

The parallel lines are

8x +15y − 34 = 0 … (1)

8x + 15y + 31 = 0 …(2)

Let d be the distancebetween the given lines.

So,

∴ The distance betweengivens parallel line is 65/17 units.

**Answer
2** :

Given:

The equation is parallelto x + 7y + 2 = 0 and at unit distance from the point (1, -1)

The equation of givenline is

x + 7y + 2 = 0 … (1)

The equation of a lineparallel to line x + 7y + 2 = 0 is given below:

x + 7y + λ =0 … (2)

The line x + 7y+ λ = 0 is at a unit distance from the point (1, − 1).

So,

1 =

λ – 6 = ± 5√2

λ = 6 + 5√2, 6 – 5√2

now, substitute thevalue of λ back in equation x + 7y + λ = 0, we get

x + 7y + 6 + 5√2 = 0 and x + 7y + 6 – 5√2

∴ The required lines:

x + 7y + 6 + 5√2 = 0 and x + 7y + 6 – 5√2

**Answer
3** :

Given:

The lines A, 2x + 3y =19 and B, 2x + 3y + 7 = 0 also a line C, 2x + 3y = 6.

Let d_{1} bethe distance between lines 2x + 3y = 19 and 2x + 3y = 6,

While d_{2} isthe distance between lines 2x + 3y + 7 = 0 and 2x + 3y = 6

Hence proved, thelines 2x + 3y = 19 and 2x + 3y + 7 = 0 are equidistant from the line 2x + 3y =6

**Answer
4** :

Given:

9x + 6y – 7 = 0 and 3x+ 2y + 6 = 0 are parallel lines

The given equations ofthe lines can be written as:

3x + 2y – 7/3 =0 … (1)

3x + 2y + 6 = 0 … (2)

Let the equation ofthe line midway between the parallel lines (1) and (2) be

3x + 2y+ λ = 0 … (3)

The distance between(1) and (3) and the distance between (2) and (3) are equal.

Now substitute thevalue of λ back in equation 3x + 2y + λ = 0, we get

3x + 2y+ 11/6 = 0

By taking LCM

18x + 12y + 11 = 0

∴ The required equationof line is 18x + 12y + 11 = 0

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