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.

If $(h,k)$ be any point on on $-2x+6y-2=0\iff x=3y-1$ we have $h=3k-1$
If $d$ is the distance between $(0,4);(3k-1,k)$
$d^2=(3k-1)^2+(k-4)^2=10k^2-14k+17=10\left(k^2-\dfrac75k\right)+17$
$=10\left(k-\dfrac7{10}\right)^2+17-10\cdot\left(\dfrac7{10}\right)^2$
$\ge17-10\cdot\left(\dfrac7{10}\right)^2$
The equality occurs if $k-\dfrac7{10}=0\iff k=\dfrac7{10}$
