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?

My only issue is that my solution doesn't account for if there was a maximum instead of a minimum on the distance equation.
Well, you can solve this problem by invoking the "first derivative test".
You got $f'(x)=0$ if $x=\frac{50}{34}$. Analogously: 


*

*$f'(x)<0$ if $x<\frac{50}{34}$

*$f'(x)>0$ if $x>\frac{50}{34}$


Thus, $f$ is decreasing on $(-\infty,\frac{50}{34})$ and increasing on $(\frac{50}{34},\infty)$. So, $f$ has a minimum at $\frac{50}{34}$.
