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The equation of the altitudes AD, BE, CF...

The equation of the altitudes AD, BE, CF of a triangle ABC are `x + y = 0, x-4y = 0 and 2x-y = 0`, respectively. lf. A = (t,-t) where t varies, then the locus of centroid of triangle ABC is (A) `y = -5x` (B) `y=x` (C) `x = -5y` (D) `x = y`

A

`y =- 5x`

B

`y = x`

C

`x =- 5y`

D

`x =- y`

Text Solution

Verified by Experts

The correct Answer is:
C
All the altitudes pass through (0,0). Hence, origin is the orthocentre `A -= (t,-t)`.
Let `B -= (4t_(1),t_(1))` and `C -= (t_(2),2t_(2))`satisfying BE and CF.
`m_(BE) = (1)/(4), m_(CF) = 2, m_(AC) =(2t_(2)+t)/(t_(2)-t)` and `m_(AB) =(t_(1)+t)/(4t_(1)-t)`
So `(1)/(4) ((2t_(2)+t)/(t_(2)-t)) =- 1` and `2((t_(1)+t)/(4t_(1)-t)) =- 1`
`rArr t_(2) =(t)/(2)` and `t_(1) =- (t)/(6)`
So `C -= ((t)/(2),t)`and `B -= (-(2)/(3)t,-(t)/(6))`
Let `G(x_(1),y_(1))` be centroid of `DeltaABC` and 't' varies.
So `x_(1)=(1)/(3) (t-(2)/(3)t+(t)/(2)) =(5t)/(18)` and
`y_(1) =(1)/(3)(-t-(t)/(6)+t) =- (t)/(18)`
Eliminating 't', we get the locus as `-x_(1) = 5y_(1)` or `x =- 5y`.
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