The value of `a` for which the image of the point `(a, a-1)` w.r.t the line mirror `3x+y=6a` is the point `(a^2 + 1, a)` is (A) 0 (B) 1 (C) 2 (D) none of these

Answer

2

Explanation

Image of point P in the AB is Q. Thus L is the midpoint of AB
`therefore L-=((a^2+a+1)/(2),(2a-1)/(2))`

Since, L lies on line AB.
`therefore 3((a^2+a+1)/(2))+(2a-1)/(2)=6a`
`rArr3a^2-7a+2=0`
`therefore a =2,(1)/(3)`
Also, `PQbotAB`.
`rArr-3xx(1)/(a^2-a+1)=-1`
`rArra^2-a-2=0`
Therefore , common value of a is 2.