We have three points $P=(2,3)$, $Q=(4,-2)$ and $R=(\alpha,0)$. We need to find the value of $\alpha$ such that:
![The Triangle[1]](https://i.stack.imgur.com/B4wXD.jpg)
Part 1: $|PR+QR|$ is maximized.
Part 2: $|PR-QR|$ is minimized.
I have the solution for both. I have understood the solution for case 1 but not for case 2.
Case 1's solution : https://i.stack.imgur.com/z18eZ.jpg
Case 2's Solution: https://i.stack.imgur.com/drhqU.jpg
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.
The minimum of $|PQ+QR|$ occurs when $R$ is collinear with $P$ and $Q$ because of the triangular inequality $|PQ+QR|\geq PQ$ then $|PQ+PR|$ is minimum when $PQ+PR|=PQ$.
To find the maximum of $|PQ-QR|$ call $Q'$ the symmetric of $Q$ wrt $x$-axis, then $QR=Q'R$ and remember that in a triangle a side is greater than the difference of the other two so $|PR-Q'R|\leq PQ'$ and the maximum for $|PR-Q'R|$ is when $R$ is collinear with $P$ and $Q'$, the intersection of the line $PQ'$ with $x$-axis.
Line $PQ'$ has equation $x+2y=8$ and substituting $y=0$ we get $x=8$ that is the solution $\alpha =8$
Hope it helps
PS
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