If α=1+∑r=16(–3)r–1 C2r–112, then the distance of the point (12,3) from the line αx–3y+1=0 is _________. [2025]
(5)
Given, α=1+∑r=16(–3)r–1 C2r–112
=1+∑r=16C2r–112(3i)2r–13i
Let t=3i, then we have
α=1+13i[C112t+C312t3+...+C1112t11]
=1+13i[(1+3i)12–(1–3i)122]
=1+13i[(–2ω2)12–(2ω)122]=1 [∵ ω3=0]
∴ Distance of (12,3) from x–3y+1=0 is
|12–3(3)+1(1)2+(–3)2|=12–3+11+3=5.