Minimising the Sum of distance of two line formed by joining some point on a line to two fixed points

Question

Consider the xy coordinate system.

Let $l$ be the line $y=-x$. Consider two points $A=(0,2)$ and $B=(2,3)$ The Question asks to find a point $C$ on the line $l$ such that $|AC|+|BC|$ is minimum.

The only method I got to tackle this problem is to use parametric form of a point on the line $l$, that is $(t,-t)$ and use calculus to minimize the distance as a function of variable $t$. I tried to find the critical points using differentiation. Unfortunately it gives me a $4$ degree equation with leading coefficient being $24$. Also since it has $4$ roots I an also getting duplicate roots Any other better approach?

Answer 1

If $A'=(-2,0)$ is the reflection of $A$ about line $l$, then $C$ is the intersection between $l$ and $A'B$. See here for a proof: https://www.cut-the-knot.org/Curriculum/Geometry/HeronsProblem.shtml.

Answer 2

Let's see. We have $A = (0, 2)$, $B = (2, 3)$, and the point $C$ (as a function of $t$) $C(t) = ( t, -t)$. The Euclidean distance between two points is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, so the sum of distance from $A$ to $C$, and the distance from $C$ to $B$, as a function of $t$ is $$\bbox{ L(t) = \sqrt{t^2 + (-t-2)^2} + \sqrt{(t - 2)^2 + (-t - 3)^2} }$$ which simplifies to $$\bbox{ L(t) = \sqrt{2 t^2 + 4 t + 4} + \sqrt{2 t^2 + 2 t + 13} }$$ Its extrema (minimum or maximum points) occur when its derivative is zero. Using the chain rule, $\frac{d \sqrt{f(x)}}{d x} = \frac{\frac{d f(x)}{d x}}{2 \sqrt{f(x)}}$, so we need to solve $$\bbox{ \frac{d L(t)}{d t} = \frac{2 t + 2}{\sqrt{2 t^2 + 4 t + 4}} + \frac{2 t + 1}{\sqrt{2 t^2 + 2 t + 13}} = 0 }$$ for $t$. Note that neither polynomial within the square roots is always positive ($\gt 0$) for all real $t$, as they're parabolas; left one has a minimum $2$ at $t = -1$, and the right one has a minimum $12.5$ at $t = -0.5$. Or, if you prefer, because neither $A$ nor $B$ is on the line, and the square roots correspond to the distance from a point on the line to $A$ and $B$, respectively, the square roots must be greater than zero for all real $t$. This means both divisors are always greater than zero, and we can multiply the equation by their product. This gives us $$\bbox{ \frac{d L(t)}{d t} = (2 t + 2) \sqrt{2 t^2 + 2 t + 13} + (2 t + 1)\sqrt{2 t^2 + 4 t + 4} = 0 }$$ and, moving the second summand to the right side, $$\bbox{ (2 t + 2) \sqrt{2 t^2 + 2 t + 13} = (- 2 t - 1) \sqrt{2 t^2 + 4 t + 4} }$$ If we restrict our solutions to those where both sides have the same sign, i.e. $(2 t + 2)$ and $(- 2 t - 1)$ have the same sign, we can square both sides. This gives us $$\bbox{ (2 t + 2)^2 ( 2 t^2 + 2 t + 13 ) = (- 2 t - 1)^2 ( 2 t^2 + 4 t + 4 ) }$$ Expanding and moving the right side back to the left side, we get $$\bbox{ 42 t^2 + 92 t + 48 = 0 }$$ This second-degree function has two roots: $t = -\frac{6}{7}$ and $t = -\frac{4}{3}$. However, remeber the restriction earlier. For $t = -\frac{4}{3}$, $2 t + 2 \lt 0$ but $-2 t - 1 \gt 0$, so $t = -\frac{4}{3}$ is not a valid solution. For $t = -\frac{6}{7}$, $2 t + 2 \gt 0$ and $-2 t - 1 \gt 0$, so $t = -\frac{6}{7}$ is a valid answer.

Therefore, the point $C$ we are looking for is $C\left(-\frac{6}{7}\right)$, i.e. $C = \left(-\frac{6}{7}, \frac{6}{7}\right)$.

As Aretino already pointed out, that is the exact same point that you get if $A^\prime$ is the reflection of $A$ about the line $y = -x$ (i.e. $A^\prime = (-2, 0)$), and you find $C$ at the intersection between lines $y = -x$ and $A^\prime B$.