Extremizing sum/difference of lengths using triangle inequality

Question

We have three points $P=(2,3)$, $Q=(4,-2)$ and $R=(\alpha,0)$. We need to find the value of $\alpha$ such that:

Part 1: $|PR+QR|$ is maximized.

Part 2: $|PR-QR|$ is minimized.

I was given the solution for both. I have understood the solution for case 1 but not for case 2.

Case 1's solution : https://imgur.com/a/ymUvM

Case 2's Solution: https://imgur.com/a/lLK9k

I have understood case 1 completely but in case 2 why was there a need to take the image of point Q? I fail to understand this.