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`.
The equation to the sides of a triangle are x-3y = 0, 4x + 3y = 5 and 3x +y=0. The line 3x – 4y = 0 passes through
The equations of the perpendicular bisectors of the sides A Ba n dA C of triangle A B C are x-y+5=0 and x+2y=0 , respectively. If the point A is (1,-2) , then find the equation of the line B Cdot
The equations of the sides of a triangle are x+y-5=0, x-y+1=0, and y-1=0. Then the coordinates of the circumcenter are
Find the area of the triangle formed by the lines y - x = 0, x + y = 0 " and " x - k = 0 .
Find the area of a triangle formed by lines 3x+y-2=0, 5x+2y-3=0 and 2x-y-3=0
The equations of two equal sides A Ba n dA C of an isosceles triangle A B C are x+y=5 and 7x-y=3 , respectively. Then the equation of side B C if a r( A B C)=5u n i t^2 is x-3y+1=0 (b) x-3y-21=0 3x+y+2=0 (d) 3x+y-12=0
In triangle A B C , the equation of the right bisectors of the sides A B and A C are x+y=0 and y-x=0 , respectively. If A-=(5,7) , then find the equation of side B Cdot
A B C is a variable triangle such that A is (1,2) and B and C lie on line y=x+lambda (where lambda is a variable). Then the locus of the orthocentre of triangle A B C is (a) (x-1)^2+y^2=4 (b) x+y=3 (c) 2x-y=0 (d) none of these
If points Aa n dB are (1, 0) and (0, 1), respectively, and point C is on the circle x^2+y^2=1 , then the locus of the orthocentre of triangle A B C is (a) x^2+y^2=4 (b) x^2+y^2-x-y=0 (c) x^2+y^2-2x-2y+1=0 (d) x^2+y^2+2x-2y+1=0
CENGAGE-COORDINATE SYSTEM-Multiple Correct Answers Type
