- 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-23.4
- 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

RD Chapter 23- The Straight Lines Ex-23.1 |
RD Chapter 23- The Straight Lines Ex-23.2 |
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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 |
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RD Chapter 23- The Straight Lines Ex-23.18 |
RD Chapter 23- The Straight Lines Ex-23.19 |

**Answer
1** :

Given: A line whichmakes an angle of 150^{o} with the x–axis and cutting off anintercept at 2

By using the formula,

The equation of a lineis y = mx + c

We know thatangle, θ = 150^{o}

The slope of the line,m = tan θ

Where, m = tan 150^{o}

= -1/ √3

Coordinate ofy–intercept is (0, 2)

The required equationof the line is y = mx + c

Now substitute thevalues, we get

y = -x/√3 + 2

√3y – 2√3 + x = 0

x + √3y = 2√3

∴ The equation of lineis x + √3y = 2√3

Find the equation of a straight line:

(i) with slope 2 and y – intercept 3;

(ii) with slope – 1/ 3 and y – intercept – 4.

(iii) with slope – 2 and intersecting the x–axis at a distance of 3 units to the left of origin.

**Answer
2** :

(i) With slope 2 and y –intercept 3

The slope is 2 and thecoordinates are (0, 3)

Now, the requiredequation of line is

y = mx + c

Substitute the values,we get

y = 2x + 3

(ii) With slope – 1/ 3 andy – intercept – 4

The slope is – 1/3 andthe coordinates are (0, – 4)

Now, the requiredequation of line is

y = mx + c

Substitute the values,we get

y = -1/3x – 4

3y + x = – 12

(iii) With slope – 2 andintersecting the x–axis at a distance of 3 units to the left of origin

The slope is – 2 andthe coordinates are (– 3, 0)

Now, the requiredequation of line is y – y_{1} = m (x – x_{1})

Substitute the values,we get

y – 0 = – 2(x + 3)

y = – 2x – 6

2x + y + 6 = 0

**Answer
3** :

There are two bisectorsof the coordinate axes.

Their inclinationswith the positive x-axis are 45^{o} and 135^{o}

The slope of thebisector is m = tan 45^{o} or m = tan 135^{o}

i.e., m = 1 or m = -1,c = 0

By using the formula,y = mx + c

Now, substitute thevalues of m and c, we get

y = x + 0

x – y = 0 or y = -x +0

x + y = 0

∴ The equation of thebisector is x ± y = 0

**Answer
4** :

Given:

The equation whichmakes an angle of tan ^{– 1}(3) with the x–axis and cuts off anintercept of 4 units on the negative direction of y–axis

By using the formula,

The equation of theline is y = mx + c

Here,angle θ = tan ^{– 1}(3)

So, tan θ = 3

The slope of the lineis, m = 3

And, Intercept in thenegative direction of y–axis is (0, -4)

The required equationof the line is y = mx + c

Now, substitute thevalues, we get

y = 3x – 4

∴ The equation of theline is y = 3x – 4.

**Answer
5** :

Given:

A line segment joining(2, – 5) and (1, 2) if it cuts off an intercept – 4 from y–axis

By using the formula,

The equation of lineis y = mx + C

It is given that, c =– 4

Slope of line joining(x_{1} – x_{2}) and (y_{1} – y_{2}),

So, Slope of linejoining (2, – 5) and (1, 2),

m = – 7

The equation of lineis y = mx + c

Now, substitute thevalues, we get

y = –7x – 4

y + 7x + 4 = 0

∴ The equation of lineis y + 7x + 4 = 0.

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