Find the point P. 
Given two points $A(1,2), B(2,3)$ and a line $L:2x+3y+5=0$. If $P$ is a variable point on $L$. Locate $P$ if
$1)\;PA+PB$ is minimum
$2)\;|PA-PB|$ is maximum

I assumed the point to be $(-1.5y-2.5,y)$ and applied distance formula for both, I took $f(P)=|PA|$ and $g(P)=|PB|$ and then,
For
$1)$
$$
g'(P)+f'(P)=0 \\
g''(P)+f''(P)>0
$$
$2)$
$$
g'(P)-f'(P)=0 \\
g''(P)-f''(P)<0
$$
It makes me hard to solve these and double check. So, I took help of wolfram alpha. Is there any other method which helps solve quickly and manually?
 A: $(x_0, y_0) = (-1, -1)$ is a point belonging to the line $L$.
Vector $\mathbf{n}=\overline{(2, 3)}$ is a normal vector to the line $L\Rightarrow$ vector $\mathbf{s} = \overline{(-3, 2)}$ is parallel to $L$.
Then, any point $P∈ L$ can be represented in a parametric form as
$$
P(t) = \left\{
\begin{aligned}
P_x &= -1-3t \\
P_y &= -1+2t
\end{aligned}
\right., \: t\in (-\infty, +\infty).
$$
$$
\begin{aligned}
PA \pm PB &= \sqrt{(1-(-1-3t))^2+(2-(-1+2t))^2} \pm \sqrt{(2-(-1-3t))^2+(3-(-1+2t))^2} = \\
&= \sqrt{(2+3t)^2+(3-2t)^2}\pm\sqrt{(3+3t)^2+(4-2t)^2} = \\
&= \sqrt{13t^2+13}\pm\sqrt{13t^2+3t+25}\\ \\
PA+PB \rightarrow \min &\Leftrightarrow \sqrt{13t^2+13}+\sqrt{13t^2+3t+25} \rightarrow \min. \\
|PA-PB| \rightarrow \max &\Leftrightarrow \left|\sqrt{13t^2+13}-\sqrt{13t^2+3t+25}\right| \rightarrow \max.
\end{aligned}
$$
Once you have a parameter value for $t$, say $t^*$, you can find $P^* = P(t^*)$.
A: $1)$ Let the reflection of $B$ in $\overleftrightarrow{L}: y=-\frac{2}{3}x-\frac{5}{3}$ be $B'$ and let $\overleftrightarrow{AB'}$ cuts the line $\overleftrightarrow{L}$ at $P$, then because the triange $\triangle BPB'$ is isoceles with $PB=P'B$, $PA+PB=PA+PB'=AB'$ and we claim that this is the minumum value. Because, for another point $Q$ on $\overleftrightarrow{L}$, $QA+QB$ will be greater than $AB'$  by triangular inequality in the triangle $\triangle AQB'$. So, how do we find this $P$? The normal of $\overleftrightarrow{L}$ is $\langle2,3\rangle$, so the line through $B(2,3)$ and $B'$ has slope $3/2$ and it is $y=\frac{3}{2}x$. It cuts $\overleftrightarrow{L}$ at $M(-\frac{10}{13},-\frac{15}{13})$. From $B+B'=2M$ we have $B'(-\frac{46}{13},-\frac{69}{13})$. The line throgh $A(1,2)$ and $B'$ is $y=\frac{95}{59}x+\frac{23}{59}$ and it cuts $\overleftrightarrow{L}$ at $P(-\frac{28}{11},-\frac{33}{31})$. And $AP+PB=AB'=\sqrt{74}$ is the minum value.
$2)$ Let the line $\overleftrightarrow{AB}$ cuts the line $\overleftrightarrow{L}$ at  $P$. Then $|AP-BP|=AB=\sqrt{(2-1)^2+(3-2)^2}=\sqrt2$ is the maximum, because for another point $Q$ on the line $\overleftrightarrow{L}$, by triangular inequality on the triangle $\triangle AQB$, we would have $|AQ-QB|<AB$.
