- 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-23.11
- 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-23.8 |
RD Chapter 23- The Straight Lines Ex-23.9 |
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 point of intersection of the following pairs of lines:

(i) 2x – y + 3 = 0 and x + y – 5 = 0

(ii) bx + ay = ab and ax + by = ab

**Answer
1** :

(i) 2x – y + 3 = 0 and x +y – 5 = 0

Given:

The equations of thelines are as follows:

2x − y + 3 = 0 … (1)

x + y − 5 = 0 … (2)

Let us find the pointof intersection of pair of lines.

By solving (1) and (2)using cross – multiplication method, we get

x = 2/3 and y = 13/3

∴ The point ofintersection is (2/3, 13/3)

(ii) bx + ay = ab and ax +by = ab

Given:

The equations of thelines are as follows:

bx + ay − ab = 0… (1)

ax + by = ab ⇒ ax + by − ab = 0 … (2)

Let us find the pointof intersection of pair of lines.

By solving (1) and (2)using cross – multiplication method, we get

∴ The point ofintersection is (ab/a+b, ab/a+b)

Find the coordinates of the vertices of a triangle, the equations ofwhose sides are: (i) x + y – 4 = 0, 2x – y + 3 0 and x – 3y + 2 = 0

(ii) y (t_{1} + t_{2}) = 2x + 2at_{1}t_{2},y (t_{2} + t_{3}) = 2x + 2at_{2}t_{3} and,y(t_{3} + t_{1}) = 2x + 2at_{1}t_{3}.

**Answer
2** :

(i) x + y – 4 = 0, 2x – y+ 3 0 and x – 3y + 2 = 0

Given:

x + y − 4 = 0, 2x − y+ 3 = 0 and x − 3y + 2 = 0

Let us find the pointof intersection of pair of lines.

x + y − 4 = 0 … (1)

2x − y + 3 = 0 … (2)

x − 3y + 2 = 0 … (3)

By solving (1) and (2)using cross – multiplication method, we get

x = 1/3, y = 11/3

Solving (1) and (3)using cross – multiplication method, we get

x = 5/2, y = 3/2

Similarly, solving (2)and (3) using cross – multiplication method, we get

x = – 7/5, y = 1/5

∴ The coordinatesof the vertices of the triangle are (1/3, 11/3), (5/2, 3/2) and (-7/5, 1/5)

(ii) y (t_{1} +t_{2}) = 2x + 2at_{1}t_{2}, y (t_{2} + t_{3})= 2x + 2at_{2}t_{3} and, y(t_{3} + t_{1})= 2x + 2at_{1}t_{3}.

Given:

y (t_{1} +t_{2}) = 2x + 2a t_{1}t_{2}, y (t_{2} + t_{3})= 2x + 2a t_{2}t_{3} and y (t_{3} + t_{1})= 2x + 2a t_{1}t_{3}

Let us find the pointof intersection of pair of lines.

2x − y (t_{1} +t_{2}) + 2a t_{1}t_{2} = 0 … (1)

2x − y (t_{2} +t_{3}) + 2a t_{2}t_{3} = 0 … (2)

2x − y (t_{3} +t_{1}) + 2a t_{1}t_{3} = 0 … (3)

By solving (1) and (2)using cross – multiplication method, we get

Solving (1) and (3)using cross – multiplication method, we get

Solving (2) and (3)using cross – multiplication method, we get

∴ The coordinatesof the vertices of the triangle are (at^{2}_{1}, 2at_{1}), (at^{2}_{2},2at_{2}) and (at^{2}_{3}, 2at_{3}).

Find the area of the triangle formed by the lines

y = m_{1}x + c_{1}, y = m_{2}x + c_{2} andx = 0

**Answer
3** :

Given:

y = m_{1}x + c_{1} …(1)

y = m_{2}x + c_{2} …(2)

x = 0 … (3)

In triangle ABC, letequations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2),we get

Solving (1) and (3):

x = 0, y = c_{1}

Thus, AB and CAintersect at A 0,c_{1}.

Similarly, solving (2)and (3):

x = 0, y = c_{2}

Thus, BC and CAintersect at C 0,c_{2}.

Find the equations of the medians of a triangle, the equations of whosesides are:

3x + 2y + 6 = 0, 2x – 5y + 4 = 0 and x – 3y – 6 = 0

**Answer
4** :

Given:

3x + 2y + 6 = 0 … (1)

2x − 5y + 4 = 0 … (2)

x − 3y − 6 = 0 … (3)

Let us assume, intriangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA,respectively.

Solving equations (1)and (2), we get

x = −2, y = 0

Thus, AB and BCintersect at B (−2, 0).

Now, solving (1) and(3), we get

x = – 6/11, y = –24/11

Thus, AB and CAintersect at A (-6/11, -24/11)

Similarly, solving (2)and (3), we get

x = −42, y = −16

Thus, BC and CAintersect at C (−42, −16).

Now, let D, E and F bethe midpoints the sides BC, CA and AB, respectively.

Then, we have:

∴ The equations ofthe medians of a triangle are: 41x – 112y – 70 = 0,

16x – 59y – 120 = 0,25x – 53y + 50 = 0

**Answer
5** :

Given:

y = √3x + 1…… (1)

y = 4 ……. (2)

y = – √3x + 2……. (3)

Let us assume intriangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA,respectively.

By solving equations(1) and (2), we get

x = √3, y = 4

Thus, AB and BCintersect at B(√3,4)

Now, solving equations(1) and (3), we get

x = 1/2√3, y = 3/2

Thus, AB and CAintersect at A (1/2√3, 3/2)

Similarly, solvingequations (2) and (3), we get

x = -2/√3, y = 4

Thus, BC and ACintersect at C (-2/√3,4)

Now, we have:

Hence proved, thegiven lines form an equilateral triangle.

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