Trigonometric EquationHard
Question
The line parallel to the x-axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0, where (a, b) ≠ (0, 0) is
Options
A.below the x-axis at a distance of 
from it
from itB.below the x-axis at a distance of 
from it
from itC.above the x-axis at a distance of from it
from itD.above the x-axis at a distance of from it
from itSolution
ax + 2by + 3b + λ(bx - 2ay - 3a) = 0
⇒ (a + bλ)x + (2b - 2aλ)y + 3b - 3λa = 0
a + bλ = 0 ⇒ λ = - a/b
⇒ ax + 2by + 3b -
(bx - 2ay - 3a) = 0
⇒ ax + 2by + 3b - ax +
= 0
so it is 3/2 units below x-axis.
⇒ (a + bλ)x + (2b - 2aλ)y + 3b - 3λa = 0
a + bλ = 0 ⇒ λ = - a/b
⇒ ax + 2by + 3b -
(bx - 2ay - 3a) = 0⇒ ax + 2by + 3b - ax +
= 0 so it is 3/2 units below x-axis.Create a free account to view solution
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