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.
 A: 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 $k-\dfrac7{10}=0\iff k=\dfrac7{10}$
A: 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.
A: 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?
