I need to find the minimal distance between a point $(x_0,y_0)$ and a line $ax+by = c$ for some constants $a,b,c ∈ ℝ$ using the Lagrange Multiplier Method.

Knowing the distance between two points, $(x_0,y_0)$ and $(x,y)$ is given by the following equation, I know that I need to minimise this equation.

$D = \sqrt{(x-x_0)^2 + (y-y_0)^2)}$

I also know that $D_{min}$ = $D^2_{min}$ , which will make it easier to use the Lagrange Method since I can "discard" the square root.

Then I let:

$f(x,y) = D^2 = (x-x_0)^2 + (y-y_0)^2$

Since this distance must be on the line $ax + by = c$ , I set my constraint:

$G(x,y) = ax + by - c = 0$


$∇f = <2x-2x_0, 2y-2y_0> ; ∇G = < a, b >$

Using the Lagrange Method for some constant $λ$, we have:

$∇f = λ∇G \Rightarrow < 2x-2x_0, 2y-2y_0 > = λ < a, b>$

Eliminating $λ$, we have:

$y = y_0 + \frac ba (x-x_0)$

Plugging this into my original constraint, $G(x,y)$, we have:

$G(x,y) = ax + b(y_0 + \frac ba (x-x_0)) = 0$

Solving for $x$, we have:

$x = \frac { c+ \frac {b^2}{a}{x_0} - by_0 }{a + \frac {b^2}a}$

Plugging back in once more, we have:

$y = \frac ba \left( \frac { c+ \frac {b^2}{a}{x_0} - by_0 }{a + \frac {b^2}a} \right) - \frac ba{x_0} + y_0$

The Hessian seems to confirm that any critical point of $f$ will be a local maximum of $f$, but the overall complexity of the point I found is making me question whether I took the right approach.

If I plug back these coordinates into $D$, I should get the minimum distance.

$D = \sqrt{\left(\frac { c+ \frac {b^2}{a}{x_0} - by_0 }{a + \frac {b^2}a}-x_0\right)^2 + \left(\frac ba \left( \frac { c+ \frac {b^2}{a}{x_0} - by_0 }{a + \frac {b^2}a} \right) - \frac ba{x_0})^2\right)}$

This seems unnecessarily complicated and wrong. Where did I go wrong (if I did)?

  • $\begingroup$ Have you tried simplifying that horrendous-looking expression that you got for $x$ before proceeding? If nothing else, your solution is undefined when $a=0$. $\endgroup$ – amd Oct 18 '18 at 23:51

Let me restart the problem from scratch.

You want to minimize $$F=(x-x_0)^2+(y-y_0)^2+\lambda (a x+b y-c)\tag 1$$ Computing the partila derivatives with respect to $x$ and $y$ and setting them equal to $0$ gives $$\frac{\partial F}{\partial x}=2(x-x_0)+a \lambda=0 \implies x=x_0-\frac 12 a \lambda\tag 2$$ $$\frac{\partial F}{\partial y}=2(y-y_0)+b \lambda=0 \implies y=y_0-\frac 12 b \lambda\tag 3$$ Now $$\frac{\partial F}{\partial \lambda}=ax+by-c=-\frac{1}{2} \lambda \left(a^2+b^2\right)+a {x_0}+b {y_0}-c\tag 4$$ $$\frac{\partial F}{\partial \lambda}=0 \implies \lambda=\frac{2 (a {x_0}+b {y_0}-c)}{a^2+b^2}$$ So, from $(2)$ and $(3)$, after simplifications $$x=\frac{b^2 {x_0}-a b {y_0}+a c}{a^2+b^2}\qquad \text{and} \qquad y=\frac{a^2 {y_0}-a b {x_0}+b c}{a^2+b^2}$$ making after simplifications

$$(x-x_0)^2+(y-y_0)^2=\frac{(ax_0+by_0-c)^2}{a^2+b^2}$$ which is the square of the well known distance of a point to a straight line.


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