Find the point P.

Question

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?

Answer 1

$1)$ 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 $|AQ-QB|<AB$.

Answer 2

$(x_0, y_0) = (-1, -1)$ 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 $P^* = P(t^*)$.