<s>[INST] Sum of distances of a variable point from two fixed points. Given that $A\equiv(4,2)$ and $B\equiv(2,4)$, find a point $P$ on the line $3x+2y+10=0$ such that $PA+PB$ is minimum.

My attempt:
In $\triangle PAB$,
$$PA+PB\ge AB$$ Hence, the minimum value should be $AB=2\sqrt 2$. But in the solution point $P$ is given as $\displaystyle(-\frac{14}{5}, -\frac{4}{5})$ from which $PA+PB$ comes out to be $10\sqrt 2$.
Where am I going wrong? Help is requested!
 [/INST] If $A(4,2), B(2,4)$ are on the same side of the line not on the same line $L=3x+2y+10=0.$
Let the image of $A$ about $L$ be $A'$, then $A'P=AP$ then $AP+BP=A'P+BP$ is minimum
when $A',P,B$ are collinear and $PA;+PB=A'B$
Image $A'(x',y')$ is obtained as
$$\frac{x'-4}{3}=\frac{y'-2}{2}=-2\frac{12+4+10}{9+4}\implies x'=-8, y'=-6$$
The equation of $A'B$ is $x-y=-2$ and the required point $P$ is thr intersection of this line with $L$, we get $P(-14/5,-4/5)$ and $AP+BP \ge A'B=10\sqrt{2}$
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