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.

We know that the closest distance between two points is the perpendicular distance between them.
So, the question becomes: For which point $P(a,b)$ on the given line and $A(0,4)$, is $AP\perp \textrm{Given line}=(-2x+6y-2=0)$?
So, slope of $AP=\dfrac{4-b}{0-a}=m_1\textrm{(say)}$ and slope of given line is $\dfrac{1}{3}=m_2\textrm{(say)}$ using the $y=mx+c$ form.
Then, by perpendicularity, $m_1m_2=(-1)$. You get one equation in terms of $a,b$ using this.
Then, $P(a,b)$ also lies on the given line. Form another equation using that and solve the two simultaneous equations.
Can you take it from here?
