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 \left|\frac{(d_1-c_1)(d_2-c_2)}{a_1b_2-a_2b_1}\right|  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  

\left|\frac{(d_1-c_1)(d_2-c_2)}{a_1b_2-a_2b_1}\right|  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  = \left|\frac{(c_1-d_1)(c_2-d_2)}{\begin{vmatrix}a_1&a_2\\b_1&b_2\end{vmatrix}}\right|

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

Therefore, Area = \left|\frac{(c_1-d_1)(c_2-d_2)}{a_1b_2-a_2b_1}\right|=\left|\frac{(d_1-c_1)(d_2-c_2)}{a_1b_2-a_2b_1}\right|

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

Therefore, \left|\frac{c_1-d_1}{\sqrt{a_1^2+b_1^2}}\right|=\left|\frac{c_2-d_2}{\sqrt{a_1^2+b_1^2}}\right|

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 \frac{2a^2}{7}  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 \frac{2a^2}{7}  sq. units.

From above solution, we know that

Area of the parallelogram = \left|\frac{(c_1-d_1)(c_2-d_2)}{a_1b_2-a_2b_1}\right|

Area of the parallelogram =  \left|\frac{(a-3a)(2a-a)}{(-9+16)}\right|=\frac{2a^2}{7}  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 \frac{\pi}{2} .

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

B = \left(\frac{mn'-ln}{l^2-m^2}, \frac{mn-ln'}{l^2-m^2}\right)

Solving equation (2) and (3), we get 

C = \left(-\frac{n'}{m+1'}, -\frac{n'}{m+1}\right)

Solving equation (3) and (4), we get 

D = \left(\frac{mn-ln'}{l^2-m^2}, \frac{mn'-ln}{l^2-m^2}\right)

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

A = \left(-\frac{n}{m+1'}, -\frac{n}{m+1}\right)  

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

Now,

m₁ = \frac{\frac{-n'}{m+1}+\frac{n}{m+1}}{\frac{-n'}{m+1}+\frac{n}{m+1}}=1

m₂ = \frac{\frac{mn'-ln}{l^2-m^2}+\frac{mn-ln'}{l^2-m^2}}{\frac{mn-ln'}{l^2-m^2}+\frac{mn'-ln}{l^2-m^2}}=-1

Therefore, 

m₁m₂ = -1

Hence proved.