JEE Main | 2014Trigonometric EquationHard
Question
Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and 5bx + 2by + d = 0 lies in the fourth quadrant and is equidistant from the two axes then
Options
A.3bc - 2ad = 0
B.3bc + 2ad = 0
C.2bc - 3ad = 0
D.2bc + 3ad = 0
Solution
Let (α, - α) be the point of intersection
∴ 4aα - 2aα + c = 0 ⇒ a = -
and 2bα - 2bα + d = 0 ⇒ α = -
⇒ 3bc = 2ad
⇒ 3bc - 2ad = 0
Alternative method :
The point of intersection will be
⇒ x =
⇒ y =
∵ Point of intersection is in fourth quadrant so x is positive and y is negative.
Also distance from axes is same
So x = - y (∵ distance from x-axis is - y as y is negative)
2ad - 2bc = - 5bc + 4ad
⇒ 3bc - 2ad = 0 .......(i)
∴ 4aα - 2aα + c = 0 ⇒ a = -
and 2bα - 2bα + d = 0 ⇒ α = -
⇒ 3bc = 2ad
⇒ 3bc - 2ad = 0
Alternative method :
The point of intersection will be
⇒ x =
⇒ y =
∵ Point of intersection is in fourth quadrant so x is positive and y is negative.
Also distance from axes is same
So x = - y (∵ distance from x-axis is - y as y is negative)
2ad - 2bc = - 5bc + 4ad
⇒ 3bc - 2ad = 0 .......(i)
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