JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Binomial Theorem

The expansion of $(x+y{)}^{2n-3}$ has $2n-2$ terms, and setting $x=y=1$ shows the sum of all its coefficients is $2^{2n-3}$. The arithmetic mean of the coefficients is therefore $\frac{2^{2n-3}}{2n-2}=16$, which holds for $n=5$ since $\frac{2^7}{8}=16$. The point is then $P(9,5)$, and its distance from $x+y=8$ is $\frac{|9+5-8|}{\sqrt{2}}=\frac{6}{\sqrt{2}}=3\sqrt{2}$.