Sum of distances from points Let point $P= (5,3)$ and a point $R$ on $y=x$ and $Q$ on the $x$ axis be such that $PQ+QR+RP$ is minimum. Then the coordinates of $Q$ are? 
 A: Reflect the point $P$ through the line $y=x$ and also through the axis $Ox$ and find the points $P'$ and $P''$. After that make the line through $P'$ and $P''$ and see the intersection with $x=y$ and the axis $Ox$. 

See that $PR=P'R$ because the line $x=y$ bisect $PP'$ and those are perpendicular. For the same reason we have $PQ=QP''$ and once $P',R,Q,P''$ lies on the same line then the sum $P'R+RQ+QP''=PR+RQ+PQ$ is minimum.
Once $P=(5,3)$ his reflection through axis $Ox$ is $P''=(5,-3)$ and his reflection through $x=y$ is $P'=(3,5)$. 
Now, make the line equation through $P'$ and $P''$ and find the intersection $Q$ between the line and the axis $Ox$. 
$$\frac{5-(-3)}{3-5}=\frac{y-5}{x-3} \Leftrightarrow y=-4x+17$$
So the point is $Q=(17/4,0)$.
A: We have a point P with coordinates (5, 3). Point R is on the line $y = x$, so it has coordinates $(a,a)$. Q is on the x-axis, so it has coordinates $(b, 0)$. Therefore, the distance PQ = $\sqrt{(5-a)^2 + (3-a)^2}$, QR = $\sqrt{(a-b)^2 + (a)^2}$, RP = $\sqrt{(b-5)^2 + (-3)^2}$.
I'm going to stop here, as this seems close to a homework problem and I don't want to solve it for you if you haven't put real thought into it yet. In the future, be sure to show some work or valid attempts at the solution - otherwise it just looks like you're getting someone to do your homework. Feel free to ask for followup or clarification.
