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

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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.

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Last Updated : 19 Jan, 2021

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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.

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.

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.

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

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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.

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.

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.

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

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.

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.