If two members of family `(2+lambda)x+(1+2lambda)y-3(1+lambda) = 0` and line x+y=0 make an equilateral triangle, the the incentre of triangle so formed is

Options

A. `((1)/(3), (1)/(3))`
B. `((7)/(6), -(5)/(6))`
C. `((5)/(6), (5)/(6))`
D. `(-(3)/(2), -(3)/(2))`

Answer

A

Explanation

Given equation of family of lines is
`(2x+y-3)+lambda(x+2y-3) =0, lambda in R.`
The family of lines is concurrent at point of intersection of 2x+y-3 = 0 and x+2y-3=0 which is A(1,1).
Two members `L_(1)" and "L_(2)` of this family and line x+y=0 form an equilateral triangle.
Let foot of perpendicular from A(1, 1) on the line x+y=0 be M(h,k).
`(h-1)/(1) = (k-1)/(1) = (-(1+1))/(2)`
`therefore M-=(h,k)-=(0,0)`
Since triangle is equilateral, AM is median, altitude and angle bisector.
`AM= sqrt((1-0)^(2)+(1-0)^(2)) = sqrt(2)`
Incentre coincides with centroid G such that AG `=(2)/(3)sqrt(2)`.
`"Slope of " AM is 1 = "tan" 45^(@)`.
`therefore G-=(1-(2sqrt(2))/(3)"cos" 45^(@), 1-(2sqrt(2))/(3)"sin" 45^(@))`
`-=((1)/(3),(1)/(3))`