# 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**

**a**_{1}x + b_{1}y + c_{1} = 0, a_{1}x + b_{1}y + d_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0, a_{2}x + b_{2}y + d_{2} = 0 is **sq. units.**

_{1}x + b

_{1}y + c

_{1}= 0, a

_{1}x + b

_{1}y + d

_{1}= 0, a

_{2}x + b

_{2}y + c

_{2}= 0, a

_{2}x + b

_{2}y + d

_{2}= 0 is

**Deduce the condition for these lines to form a rhombus.**

**Solution:**

Given:

The given lines are

a

_{1}x + b_{1}y + c_{1}= 0 —(equation-1)a

_{1}x + b_{1}y + d_{1}= 0 —(equation-2)a

_{2}x + b_{2}y + c_{2}= 0 —(equation-3)a

_{2}x + b_{2}y + d_{2}= 0 —(equation-4)Let us prove, the area of the parallelogram formed by the lines a

_{1}x + b_{1}y + c_{1}= 0, a_{1}x + b_{1}y + d_{1}= 0, a_{2}x + b_{2}y + c_{2}= 0, a_{2}x + b_{2}y + d_{2}= 0 issq. units.

The area of the parallelogram formed by the lines a

_{1}x + b_{1}y + c_{1}= 0, a_{1}x + b_{1}y + d_{1}= 0, a_{2}x + b_{2}y + c_{2}= 0 and a_{2}x + b_{2}y + d_{2}= 0 is given below:Area =

Since, \begin{vmatrix}a_1&a_2\\b_1&b_2\end{vmatrix} = a

_{1}b_{2}– a_{2}b_{1}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 m

_{1}and m_{2}be the slope of AC and BD.Now,

m

_{1}=m

_{2}=Therefore,

m

_{1}m_{2}= -1Hence proved.