Find a point on the line $-2x + 6y - 2 = 0$ that is closest to the point $(0,4).$ I understand that you would use the distance formula here, but I'm confused as to how you calculate x and y after that. Thanks.

hint: the point $P = (x,y)$, with $x = 3y-1$ is on the line, and the distance from it to the point $Q = (0,4)$ is: $PQ = \sqrt{(3y-1-0)^2+(y-4)^2}=f(y)$. You can now find $f'(y) = 0$ to find critical points and use the second derivative test to check the minimum value which is the shortest distance.
