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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $\frac{50}{34}$.

Answer 4

Use the fact that the distance from a point $(x_0, y_0)$ to a line $ax + by + c = 0$ is given by $$\frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$$ Plugging in the numbers in question, we get $$\frac{|-4(1) + 1(3) + 3|}{\sqrt{1^2 + 4^2}} = \frac{2\sqrt{17}}{17}.$$

If you want a proof of the formula, check here.

Answer 5

A simple, Linear Algebra approach. First,represent the intercept, the point and the direction of the line as vectors.

$p=<0,-3>,q=<1,3>,v=<1,4>$

Create the vector projection of the line between $p$ and $q$ onto $\vec{v}$

$\vec{u}=\vec{q}-\vec{p}, w=\frac{\vec{u}.\vec{v}}{\vec{v}.{\vec{v}}}\vec{v}$

Giving $\vec{u}=<1,6>,\vec{w}=<\frac{25}{17},\frac{100}{17}>$ then

the closest point is given by $r=\vec{p}+\vec{w} = <1,-3>+<\frac{25}{17},\frac{49}{17}>$

Answer 6

If you have the cartesian equation of the straight line $L$, you have a normal vector $\vec n$ to $L$. The projection $H$ of a point $A$ is the point $H$ of the line passing by $A$ directed by this normal vector, $A+t\,\vec n, \enspace t\in\mathbf R$, which satisfies the equation of $L$.

Here, the equation is $\;4x-y-3=0$, $\;\vec n=(4,-1)$, $A=(1,3)$, $H=(1+4t,3-t)$, hence the parameter $t$ is determined by the equation $$3-t=4(1+4t)-3\iff t=\frac 2{17}.$$

Answer 7

The closest point on a line is where a circle with an origin at the point touches the line. To find the radius of the circle, solve the equations for a circle and a line where there is a single solution.

Answer 8

Since the curve is linear, we can use linear algebra just fine.

Notice that $(0,-3)$ is on the line by putting $x=0$ to get $L(0)=-3$. Translate everything on the plane so that the line passes through the origin, giving a new line $L'(x)=4x$. A choice of this translation vector is $(0,3)$ as found above. The point $(1,3)$ is translated to $(1,6)$.

Now project the point $(1,6)$ onto $L'$ by using dot product. Explicitly, you evaluate the following$$\frac{\langle(1,6),(1,4)\rangle}{1^2+4^2}(1,4)$$ to obtain the vector $$(\frac{25}{17},\frac{100}{17}),$$ which is the point on $L'$ being closest to $(1,6)$. Translate back to the original position, giving $$(\frac{25}{17},\frac{100}{17})-(0,3)=(\frac{25}{17},\frac{49}{17})$$ being the point on $L$ closest to $(1,3)$.