ABCD is a square with side AB = 2. A point P moves such that its distance from A equals its distance from the line BD. The locus of P meets the line AC at `T_1` and the line through A parallel to BD at `T_2` and `T_3`. The area of the triangle `T_1 T_2 T_3` is :

Options

A. 1
B. 2
C. 3
D. 4

Answer

B

Explanation

As it starts from the mean position,
`therefore x = A "sin" omega t =A "sin" ((2pi t)/(T))`
`therefore (A)/(2)=A "sin" ((2pi t_(1))/(T)) therefore (1)/(2)="sin" (pi)/(6)="sin" (2pi t_(1))/(T)`
`therefore (pi)/(6)=(2pit_(1))/(T) therefore t_(1)=(T)/(12)`
and it will go form 0 to C in time `(T)/(4)`
`therefore " Time " (t_(2))` required to go from B to C
`= (T)/(4)-(T)/(12)=(T)/(6)`
`therefore (t_(2))/(t_(1))=(T//6)/(T//12)=(12)/(6)=2`