Is there a better way to find the closest point on a line?

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.


You can make the problem simpler since minimizing the distance is the same as minimizing the square of the distance.

Then $$D^2=(x-1)^2+(4x-3-3)^2=17 x^2-50 x+37$$ $$\frac{d(D^2)}{dx}=34 x-50$$ $$\frac{d^2(D^2)}{dx^2}=34 > 0$$

So the first derivative cancels if $x=\frac{25}{17}$ to which cooresponds $y=\frac{49}{17}$, $D^2=\frac{4}{17}$. The second derivative test confirms that this is a minimum.


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 $\frac{50}{34}$.

Tags:

Optimization

Calculus

Derivatives

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