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

• Last Updated : 19 Jan, 2021

### 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.

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