Is there a better way to find the closest point on a line? I'm given a question that asks: "Find the point on $L(x) = 4x-3$ that is closest to the point $(1,3)$."
My best guess was to find the derivative of the distances and set it equal to zero and solve to attempt to find a minimum. I come up with the derivative being$$f'(x)=\frac{1}{2}(17x^2-50x+37)^{-\frac{1}{2}}(34x-50)$$ 
And solving for x I end up with $50/34$ or about 1.4705. Now all I have to do is just plug that into the original linear equation. And when I graphed it out in desmos that appears to solve the problem correctly. My only issue is that my solution doesn't account for if there was a maximum instead of a minimum on the distance equation. Is there a more correct solution to this problem?

Since your curve is a straight line with a slope of $4$, a perpendicular to this line will have a slope of $-1/4$.
Using point-slope form, a line with a slope of $-1/4$ going through $(1,3)$ will have the following equation:
$$y-3 = (-1/4)(x-1)$$
Solving for the intersection of this line and the original line will give you your answer.
EDIT: The comments noted that I jumped in using the perpendicular as the shortest distance without explaining why it was so.  Here's a brief argument which can be worked out with more rigor if desired.  Take the $x-y$ axes, which are perpendicular.  Take the point $(0,5)$ which is $5$ units away from the origin, which lies on the $x$-axis and is also on the perpendicular from that point to the $x$ axis.  If we go in either direction along the $x$ axis, the distance from $(0,5)$ to this point on the $x$ axis will increase, because now it's going along the hypotenuse of a right triangle defined by the origin, the point $(0,5)$, and the point of intersection, and this has to be longer than $5$ units.  So, the perpendicular is the minimum distance.
