Answer
$(2,3)$.
Work Step by Step
Step 1. Let the point be $P(x,y)$, we have $y=x+1$ since it is on the line.
Step 2. Its distance $(d)$ to $(4,1)$ satisfies $d^2=(x-4)^2+(y-1)^2=(x-4)^2+(x)^2$
Step 3. Let $f(x)=(x-4)^2+(x)^2=2x^2-8x+16$
Step 4. The minimum of $f(x)$ happens when $x=-\frac{-8}{2(2)}=2$, thus we have $P(2,3)$.
