the question
Let $a,b,d,e$ be real numbers with $a>d$ and $c=a^2+b^2, f=d^2+e^2, m=\frac{b-e}{a-d} , n=\frac{bd-ae}{a-d}$. Show that if $x\in [-a,-d] $ and $y=mx+n$, then
$$E(x,y)=\sqrt{x^2+y^2+2ax+2by+c}+\sqrt{x^2+y^2+2dx+2ey+f}$$
does not depend on $x$.
The idea
We can write $E(x,y)= \sqrt{x^2+y^2+2ax+2by+a^2+b^2}+\sqrt{x^2+y^2+2dx+2ey+d^2+e^2}=\sqrt{(a+x)^2+(b+y)^2}+\sqrt{(d+x)^2+(e+y)^2}$
These got me thinking of the Minkowski inequality, I tried applying it, but I don't think it would help us show that the expression doesn't depend on x.(maybe the case of equality would...)
I also tried writing the variables in different forms using the given equalities, but honestly, I got to nothing useful and I got lost in all these variables.
Im pretty sure these are just put to distract us from the solution.
Hope one of you can help me! Thank you!
Hint: Let us define the three points: $P(a,b)$, $Q(d,e)$ and $R(-x,-y)$. The expression is equal to the sum of 2 distances: $$\mathcal E = PR + QR$$
And it’s not difficult to prove that $P, Q,R$ lie on the same line by proving that this equation holds true (the LHS is the slope of $PR$, the RHS is the slope of $PQ$. The 3 points are colinear if and only if the two slopes are equal): $$\frac{b+y}{ a+x} = \frac{b-e}{a-d}$$ By consequence $$\mathcal E = PQ = \text {const}$$