Class 11 RD Sharma Solutions- Chapter 23 The Straight Lines- Exercise 23.17

Question 1. Prove that the area of the parallelogram formed by the lines

a₁x + b₁y + c₁ = 0, a₁x + b₁y + d₁ = 0, a₂x + b₂y + c₂ = 0, a₂x + b₂y + d₂ = 0 is sq. units.

Deduce the condition for these lines to form a rhombus.

Solution:

Given:

The given lines are

a₁x + b₁y + c₁ = 0 —(equation-1)

a₁x + b₁y + d₁ = 0 —(equation-2)

a₂x + b₂y + c₂ = 0 —(equation-3)

a₂x + b₂y + d₂ = 0 —(equation-4)

Let us prove, the area of the parallelogram formed by the lines a₁x + b₁y + c₁ = 0, a₁x + b₁y + d₁ = 0, a₂x + b₂y + c₂ = 0, a₂x + b₂y + d₂ = 0 is  

 sq. units.

The area of the parallelogram formed by the lines a₁x + b₁y + c₁ = 0, a₁x + b₁y + d₁ = 0, a₂x + b₂y + c₂ = 0 and a₂x + b₂y + d₂ = 0 is given below:

Area  = 

Since, \begin{vmatrix}a_1&a_2\\b_1&b_2\end{vmatrix} = a₁b₂ – a₂b₁

Therefore, Area = 

If the given parallelogram is a rhombus, then the distance between the pair of parallel lines is equal.

Therefore, 

Hence proved.

Question 2. Prove that the area of the parallelogram formed by the lines 3x – 4y + a = 0, 3x – 4y + 3a = 0, 4x – 3y – a = 0 and 4x – 3y – 2a = 0 is  sq. units.

Solution:

Given:

The given lines are

3x − 4y + a = 0 —(equation-1)

3x − 4y + 3a = 0 —(equation-2)

4x − 3y − a = 0 —(equation-3)

4x − 3y − 2a = 0 —(equation-4)

We have to prove, the area of the parallelogram formed by the lines 3x – 4y + a = 0, 3x – 4y + 3a = 0, 4x – 3y – a = 0 and 4x – 3y – 2a = 0 is  sq. units.

From above solution, we know that

Area of the parallelogram = 

Area of the parallelogram =  sq. units

Hence proved.

Question 3. Show that the diagonals of the parallelogram whose sides are lx + my + n = 0, lx + my + n’ = 0, mx + ly + n = 0 and mx + ly + n’ = 0 include an angle π/2.

Solution:

Given:

The given lines are

lx + my + n = 0 —(equation-1)

mx + ly + n’ = 0 —(equation-2)

lx + my + n’ = 0 —(equation-3)

mx + ly + n = 0 —(equation-4)

We have to prove, the diagonals of the parallelogram whose sides are lx + my + n = 0, lx + my + n’ = 0, mx + ly + n = 0 and mx + ly + n’ = 0 include an angle .

By solving equation (1) and (2), we will get

B = 

Solving equation (2) and (3), we get 

C = 

Solving equation (3) and (4), we get 

D = 

Solving equation (1) and (4), we get

A = 

Let m₁ and m₂ be the slope of AC and BD.

Now,

m₁ = 

m₂ = 

Therefore, 

m₁m₂ = -1

Hence proved.