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
Answers
Answered by
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Let ABCD is a l l gm in which eq. of AB is
3x+4y=7a , DC is 3x+4y =7b , AD is 4x+3y=7c
and BC is 4x+3y = 7d.
By solving the above equations the coordinates of A(4c-3a,4a-3c) , B (4d-3a,4a-3d)
C (4d-3b,4b-3d) and D (4c-3b,4b-3c).
Let perpendicular drawn from D to AB is DT.
DT =(12c-9b+16b-12c-7a)/(3^2+4^2)^1/2 units
=7(b-a)/5 units.
base AB=[(4c-3a-4d+3a)^2+(4a-3c-4a+3d)^2]^1/2
AB=[(4c-4d)^2+(-3c+3d)^2]^1/2
AB=[(4)^2.(c-d)^2 +(-3)^2.(c-d)^2]^1/2.
AB= [25.(c-d)^2]^1/2 = 5.(c-d)
Area of l l gram ABCD = base × height
= AB × DT
=5.(c-d)×7.(b-a)/5 units^2
=7.(b-a).(c-d) sq.units. , Answer
3x+4y=7a , DC is 3x+4y =7b , AD is 4x+3y=7c
and BC is 4x+3y = 7d.
By solving the above equations the coordinates of A(4c-3a,4a-3c) , B (4d-3a,4a-3d)
C (4d-3b,4b-3d) and D (4c-3b,4b-3c).
Let perpendicular drawn from D to AB is DT.
DT =(12c-9b+16b-12c-7a)/(3^2+4^2)^1/2 units
=7(b-a)/5 units.
base AB=[(4c-3a-4d+3a)^2+(4a-3c-4a+3d)^2]^1/2
AB=[(4c-4d)^2+(-3c+3d)^2]^1/2
AB=[(4)^2.(c-d)^2 +(-3)^2.(c-d)^2]^1/2.
AB= [25.(c-d)^2]^1/2 = 5.(c-d)
Area of l l gram ABCD = base × height
= AB × DT
=5.(c-d)×7.(b-a)/5 units^2
=7.(b-a).(c-d) sq.units. , Answer
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