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 β(equation-1)
a1x + b1y + d1 = 0 β(equation-2)
a2x + b2y + c2 = 0 β(equation-3)
a2x + b2y + d2 = 0 β(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 β 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 β 4y + a = 0, 3x β 4y + 3a = 0, 4x β 3y β a = 0 and 4x β 3y β 2a = 0 is sq. units.
Solution:
Given:
The given lines are
3x β 4y + a = 0 β(equation-1)
3x β 4y + 3a = 0 β(equation-2)
4x β 3y β a = 0 β(equation-3)
4x β 3y β 2a = 0 β(equation-4)
We have to prove, the area of the parallelogram formed by the lines 3x β 4y + a = 0, 3x β 4y + 3a = 0, 4x β 3y β a = 0 and 4x β 3y β 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 + nβ = 0, mx + ly + n = 0 and mx + ly + nβ = 0 include an angle Ο/2.
Solution:
Given:
The given lines are
lx + my + n = 0 β(equation-1)
mx + ly + nβ = 0 β(equation-2)
lx + my + nβ = 0 β(equation-3)
mx + ly + n = 0 β(equation-4)
We have to prove, the diagonals of the parallelogram whose sides are lx + my + n = 0, lx + my + nβ = 0, mx + ly + n = 0 and mx + ly + nβ = 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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