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Question: Let P be point $(5,3)$ on the coordinate plane. Let Q and R lie on $x$-axis and line $y=x$， respectively. Find Q if

$$PQ+QR+RP$$

is minimum.

I tried this question by taking any general point R on the line y=x and using triangular inequality, which gives me $PQ+QR+RP>2RQ$ and equality tolds when P, Q and R are collinear. The above logic leads me nowhere.

My teacher told me that to do this question, we take images of fixed point in the given lines and join the images. The points of intersection obtained by this line and the given lines give us Q and R for minimum value of $PQ+QR+RP$. I verified it for a few other random cases where it seemed to work.

Though I understand the procedure, I am still not able to figure why this works. Any help would be appreciated.

Quanto
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Shubhav Jain
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1 Answers1

4

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Let $P_R(3,5)$ be the mirror image of the given point $P(5,3)$ with respect to the line $y=x$ and $P_Q(5,-3)$ be the mirror image of $P$ with respect to the $x$-axis. Then, for the point $R$ on the line $y=x$ and the point $Q$ on the $x$-axis, we have,

$$PR = RP_R, \>\>\>\>\>PQ = QP_Q$$

and the perimeter of the triangle $PQR$ is equal to

$$P_RR + RQ + Q P_Q$$

which is a path between the point $P_R$ and $P_Q$. It is known that the shortest path between any two points is a straight line. Therefore, for the triangle $PQR$ to have the minimum perimeter, the point $Q$ has to be the intersection of the line $P_QP_R$ with the $x$-axis, which yields the location of the point $Q(\frac{17}4,0)$.

Quanto
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