- 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-23.6
- 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 |
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-23.6 |
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 |

Find the equation of the straight lines passing through the following pair of points:

(i) (0, 0) and (2, -2)

(ii) (a, b) and (a + c sin α, b + c cos α)

**Answer
1** :

(i) (0, 0) and (2, -2)

Given:

(x_{1}, y_{1})= (0, 0), (x_{2}, y_{2}) = (2, -2)

The equation of theline passing through the two points (0, 0) and (2, −2) is

By using the formula,

y = – x

∴ The equation of lineis y = -x

(ii) (a, b) and (a + c sinα, b + c cos α)

Given:

(x_{1}, y_{1})= (a, b), (x_{2}, y_{2}) = (a + c sin α, b + c cos α)

So, the equation ofthe line passing through the two points (0, 0) and (2, −2) is

By using the formula,

y – b = cot α (x – a)

∴ The equation of lineis y – b = cot α (x – a)

Find the equations to the sides of the triangles the coordinates of whose angular points are respectively:

(i) (1, 4), (2, -3) and (-1, -2)

(ii) (0, 1), (2, 0) and (-1, -2)

**Answer
2** :

(i) (1, 4), (2, -3) and(-1, -2)

Given:

Points A (1, 4), B (2,-3) and C (-1, -2).

Let us assume,

m_{1,} m_{2,} andm_{3} be the slope of the sides AB, BC and CA, respectively.

So,

The equation of theline passing through the two points (x_{1}, y_{1})and (x_{2}, y_{2}).

Then,

m_{1 =} -7, m_{2} =-1/3 and m_{3} = 3

So, the equation ofthe sides AB, BC and CA are

By using the formula,

y – y_{1}= m(x – x_{1})

=> y – 4 = -7 (x –1)

y – 4 = -7x + 7

y + 7x = 11,

=> y + 3 = (-1/3)(x – 2)

3y + 9 = -x + 2

3y + x = – 7

x + 3y + 7 = 0 and

=> y + 2 = 3(x+1)

y + 2 = 3x + 3

y – 3x = 1

So, we get

y + 7x =11, x+ 3y + 7=0 and y – 3x = 1

∴ The equation of sidesare y + 7x =11, x+ 3y + 7 =0 and y – 3x = 1

(ii) (0, 1), (2, 0) and(-1, -2)

Given:

Points A (0, 1), B (2,0) and C (-1, -2).

Let us assume,

m_{1,} m_{2,} andm_{3} be the slope of the sides AB, BC and CA, respectively.

So,

The equation of theline passing through the two points (x_{1}, y_{1})and (x_{2}, y_{2}).

Then,

m_{1} =-1/2, m_{2} = -2/3 and m_{3}= 3

So, the equation ofthe sides AB, BC and CA are

By using the formula,

y – y_{1}= m(x – x_{1})

=> y – 1 = (-1/2)(x – 0)

2y – 2 = -x

x + 2y = 2

=> y – 0 = (-2/3)(x – 2)

3y = -2x + 4

2x – 3y = 4

=> y + 2 = 3(x+1)

y + 2 = 3x + 3

y – 3x = 1

So, we get

x + 2y = 2, 2x – 3y =4and y – 3x = 1

∴ The equation of sidesare x + 2y = 2, 2x – 3y =4 and y – 3x = 1

**Answer
3** :

Given:

A (−1, 6), B (−3, −9)and C (5, −8) be the coordinates of the given triangle.

Let us assume: D, E,and F be midpoints of BC, CA and AB, respectively. So, the coordinates of D, Eand F are

Median AD passesthrough A (-1, 6) and D (1, -17/2)

So, by using theformula,

4y – 24 = -29x – 29

29x + 4y + 5 = 0

Similarly, Median BEpasses through B (-3,-9) and E (2,-1)

So, by using theformula,

5y + 45 = 8x + 24

8x – 5y – 21=0

Similarly, Median CFpasses through C (5,-8) and F(-2,-3/2)

So, by using theformula,

-14y – 112 = 13x – 65

13x + 14y + 47 = 0

∴ The equation of linesare: 29x + 4y + 5 = 0, 8x – 5y – 21=0 and 13x + 14y + 47 = 0

**Answer
4** :

Given:

The rectangle formedby the lines x = a, x = a’, y = b and y = b’

It is clear that, thevertices of the rectangle are A (a, b), B (a’, b), C (a’, b’) and D (a, b’) .

The diagonal passingthrough A (a, b) and C (a’, b’) is

By using the formula,

(a’ – a)y – b(a’ – a)= (b’ – b)x – a(b’ – b)

(a’ – a) – (b’ – b)x =ba’ – ab’

Similarly, thediagonal passing through B (a’, b) and D (a, b’) is

By using the formula,

(a’ – a)y – b(a – a’)= (b’ – b)x – a’ (b’ – b)

(a’ – a) + (b’ – b)x =a’b’ – ab

∴ The equation ofdiagonals are y(a’ – a) – x(b’ – b) = a’b – ab’ and

y(a’ – a) + x(b’ – b)= a’b’ – ab

**Answer
5** :

Given:

The vertices oftriangle ABC are A (-1, -2), B(0, 1) and C(2, 0).

Let us find theequation of median through A.

So, the equation of BCis

By using the formula,

x + 2y – 2 = 0

Let D be the midpointof median AD,

4y + 8 = 5x + 5

5x – 4y – 3 = 0

∴ The equation of lineBC is x + 2y – 2 = 0

The equation of medianis 5x – 4y – 3 = 0

Name:

Email:

Copyright 2017, All Rights Reserved. A Product Design BY CoreNet Web Technology