Question

$\textbf{\huge{Hello!!!}}$

The area of the parallelogram formed by the lines 3x-4y+1=0 , 3x-4y+3=0 , 4x-3y-1=0 and 4x -3y-2=0, is:

(a)1
--- sq units
7

(b)2
--- sq units
7

(c)3
--- sq units
7

(d)4
--- sq units
7


$\fcolorbox{black}{red}{no spam}$

Answer 1

Let the parallelogram  ABCD  as shown is  figure

Now,

We have  to  find  base

Find the coordinates   of B

Solution of  line  AB  and  BC will be  coordinates  of  B

Equations,

3x - 4y + 1 = 0

4x - 3y - 1 = 0

on balancing  equations;-

12x - 16 y + 4 = 0

12x - 9y - 3 = 0

Substracting,

-7y + 7 = 0

y = 1

putting  y = 1 in first  equation,

x = 1

Coordinates  of B = (1,1)

Similarly,

Find the Coordinates  of A ;-

Solution  of  line  AB and AD  will be  coordinates of  A  ,

equations,

3x - 4y + 1 = 0

4x - 3y - 2 = 0

on Balancing eqautions

12x - 16y + 4 = 0

12x  - 9y - 6 = 0

Substracting ,

-7y + 10 = 0

y = 10/7

putting  y = 10/3 in    first  equation  

x = 11/7

Hence, Coordinates  of  A  = (11/7, 10/7)

Find  the distance AB;

$AB=\sqrt{(\frac{10}{7}-1)^{2}+(\\\frac{11}{7}-1)^{2}}$

AB= 5/7

Find the height of   parallelogram;
Distance between 3x-4y+1= 0 and 3x-4y + 3= 0
h= (c1-c2)/√(a²+b²)
h= (3-1)/√(3²+4²)
h = 2/5

Now,

Area of   parallelogram = height × base

                                       = 2/5*5/7 = 2/7 sq units