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!

If you draw the graph, you will understand Michael's argument. To find the coordinate of $A^\prime$ quickly, you can use the formula for the distance from a point $(a,b)$ to line $Ax+By+C=0$ as follows:
$$d=\frac{|Aa+Bb+C|}{\sqrt{A^2+B^2}}.$$
Suppose the coordinate of $A^\prime$ is $(a, b)$, then
$$\frac{-(3a+2b+10)}{\sqrt{3^2+2^2}}=\frac{3(4)+2(2)+10}{\sqrt{3^2+2^2}}.$$
The reason of adding a negative sign to the left-hand side is because $A^\prime$ lies under the line $3x+2y+10=0$.
To solve $a$ and $b$, you need to add another equation from the fact that $AA^\prime$ is perpendicular to the line $3x+2y+10=0$.
$$\frac{b-2}{a-4}\cdot \frac{-3}{2}=-1.$$
