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!
On
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 $3x+2y+10=0$. $$\frac{b-2}{a-4}\cdot \frac{-3}{2}=-1.$$
On
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 $AP+BP \ge A'B=10\sqrt{2}$
Let $A'$ be symmetric to $A$ respect to $l:3x+2y+10=0$.
Thus, $BA'\cap l=\{P\}$.
I got, since $AA'\perp l,$ we obtain: $$AA'\cap L=\{(-2,-2)\},$$ $$A'(-8,-6),$$ the equation of $BA'$ it's $y=x+2$ and $$P\left(-\frac{14}{5},-\frac{4}{5}\right).$$