Q.

If α=1+r=16(3)r1 C2r112, then the distance of the point (12,3) from the line αx3y+1=0 is _________.         [2025]


Ans.

(5)

Given, α=1+r=16(3)r1 C2r112

                    =1+r=16C2r112(3i)2r13i

Let t=3i, then we have

α=1+13i[C112t+C312t3+...+C1112t11]

        =1+13i[(1+3i)12(13i)122]

       =1+13i[(2ω2)12(2ω)122]=1         [ ω3=0]

   Distance of (12,3) from x3y+1=0 is

          |123(3)+1(1)2+(3)2|=123+11+3=5.