# Minimum Distance from an Arbitrary Point in Two Dimensions and a Line Using Lagrange Multipliers

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:

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

Then:

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

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)?

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

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.