Coordinate of point $C$ on the line So that Perimeter of a $\triangle ABC$ is minimum

Question

Given $2$ points $A(-2,0)$ and $B(0,4)$ and a line $y=x.$ Find the coordinate of a

point $C$ on the line So that Perimeter of a $\triangle ABC$ is minimum

$\bf{My\; Try::}$ Let Coordinate of Point $P(x,y)$

Here We have to Minimize $AB+BC+CA = 2\sqrt{5}+\sqrt{(x+2)^2+y^2}+\sqrt{x^2+(y-4)^2}$

So We have to Minimize $CA+CB$

So

$$\min\left(\sqrt{(x+2)^2+y^2}+\sqrt{x^2+(y-4)^2}\right)$$

So Using Minkowski Inequality::

$$\sqrt{(x+2)^2+y^2}+\sqrt{x^2+(y-4)^2}\geq \sqrt{(x+2-x)^2+(y-y+4)^2} = 2\sqrt{5}$$

and equality hold when $$\frac{x+2}{-x} = \frac{y}{-(y-4)}\Rightarrow 2x+4=y$$

So solving $2x+4=y$ and $y=x\;,$ We get $(x,y) = (-4,-4)$

Although I have used Minkowski Inequality, But i did not understand how can i solve it Using Image of point $A$ or $B$ with respect to line $y=x$

plz explain me using Figure, Thanks

Answer 1

We know light travels along path of least time. Using this, you get $C = (0,0)$, which is not very hard to justify.

The Length $AB$ is fixed. So we seek to minimise $AC + CB$, where Fermat's principle tells us that its minimum would be along reflected path.

So we take reflection of $A$ about $y = x$, and find distance $A'B$

So the perimeter will be P = $A'B + AB$

Answer 2

By Fermat principle minimum length/time taken when angle of incidence equals that of reflection.

The plane mirror is defined by $x=y $. It mirrors given points $A,B$ to $(a,b)$, as $ ((0,-2),(4,0)). $

Now connect diagonal points $(A,b)$ and $(a,B)$ by a straight line to find the required concurrent point on mirror line $x=y$ as the origin $(0,0)$.