Given two points $A(-2,0)$ and $B(0,4)$ then find coordinate of point $P$ lying on the line $2x-3y=9$ so that perimeter of triangle $APB$ is least.
Doing it by traditional calculus is making calculations very complicated. Is there smart geometrical way to solve this?
Well, it was easy to say what might help, but a bit less simple to find a way to avoid unpleasant radicals, which are essentially the same ones you would get using path-minimization in calculus. Here is a geometric argument that dodges them.
The big problem with a direct approach is that the lengths of $ \ AP \ $ and $ \ BP \ $ are going to be unequal, and a number of approaches still force us to determine them. Instead, extend a line parallel to $ \ 2x \ - \ 3y \ = \ 9 \ $ that passes through $\ A \ (-2, \ 0 ) \ $ : this is $ \ y \ = \ \frac{2}{3} x \ + \ \frac{4}{3} \ $ . This line intersects $ \ BP \ $ at a point we'll call $ \ Q \ $ . If we designate $ \ P \ ( X, \ Y) \ = \ (X, \ \frac{2}{3}X \ - \ 3) \ $ , then the line containing $ \ BP \ $ is $ \ y \ - \ 4 \ = \ M x \ $ , which has slope
$$ M \ = \ \frac{4 \ - \ [ \ \frac{2}{3}X \ - \ 3 \ ]}{0 \ - \ X \ } \ \ = \ \ \frac{2}{3} \ - \ \frac{7}{X} \ \ . $$
The coordinates of the intersection point $ \ Q \ $ are then found from equating
$$ \frac{2}{3} x_Q \ + \ \frac{4}{3} \ = \ \left( \ \frac{2}{3} \ - \ \frac{7}{X} \ \right) x_Q \ + \ 4 \ \ \Rightarrow \ \ x_Q \ = \ \frac{8X}{21} \ \ , \ \ y_Q \ = \ \frac{16X \ + \ 84}{63} \ \ . $$
(I am sparing you a certain amount of algebra in this.)
Here is where we apply the "optical law" that we minimize the length of $ \ AP \ $ plus $ \ PB \ $ by equating the angles these segments make to the normal line to $ \ 2x \ - \ 3y \ = \ 9 \ $ . The normal line has slope $ \ -\frac{3}{2} \ $ and is the perpendicular bisector of $ \ AQ \ $ when $ \Delta APQ \ $ is isoceles ; then the segment $ \ PM \ $ bisects $ \angle APQ \ $ , and so also $ \ \angle APB \ $ . The midpoint $ \ M \ $ of segment $ \ AQ \ $ is
$$ \ x_M \ = \ \frac{\frac{8X}{21} \ + \ (-2)}{2} \ = \ \frac{4X}{21} \ - \ 1 \ \ , \ \ y_M \ = \ \frac{8X \ + \ 42}{63} $$
and lies on the normal line through $ \ P \ $ , which is
$$ y \ - \ ( \ \frac{2}{3}X \ - \ 3 \ ) \ = \ -\frac{3}{2} \ (x \ - \ X ) \ \ . $$
Omitting some further algebra, after inserting
$$ y_ M \ - \ ( \ \frac{2}{3}X \ - \ 3 \ ) \ = \ -\frac{3}{2} \ (x_ M \ - \ X ) \ \ , $$
we solve for $ \ X \ $ to obtain
$$ X \ = \ \frac{21}{17} \ \ , \ \ Y \ = \ -\frac{111}{51} \ \ . $$
The graph below shows the geometrical situation.
There is no symmetry in the arrangement, so there's no good reason to expect the coordinates to come out "nicely", or that there is any "easy" method.
EDIT (added 12 March) --
Apart from "verifying" this graphically by plotting the function for the total length $ \ AP \ + \ BP \ $ , we can confirm the exact value for $ \ X \ $ by differentiating implicitly the total length
$$ s \ = \ s_1 \ + \ s_2 \ = \ \sqrt{(X \ + \ 2)^2 + (\frac{2}{3}Y \ - \ 3)^2} \ + \ \sqrt{X^2 \ + \ ( \ [ \ \frac{2}{3}X \ - \ 3 \ ] \ - \ 4 \ )^2} $$
to produce
$$ \frac{ds}{dX} \ = \ 0 \ = \ \frac{ds_1}{dX} \ + \ \frac{ds_2}{dX} \ \ \Rightarrow \ \ \frac{ds_1}{dX} \ = \ -\frac{ds_2}{dX} \ \ ;$$
squaring the individual terms $ \ s_1 \ $ and $ \ s_2 \ $ to remove the square-roots and then differentiating, we have
$$ 2 \ s_1 \ \frac{ds_1}{dX} \ = \ 2 \ (X \ + \ 2) \ + \ 2 \ ( \frac{2}{3}X \ - \ 3) \ \cdot \frac{2}{3} \ \ , $$ $$ 2 \ s_2 \ \frac{ds_2}{dX} \ = \ 2 \ X \ + \ 2 \ ( \frac{2}{3}X \ - \ 7) \ \cdot \frac{2}{3} $$
$$ \Rightarrow \ \ \frac{(X \ + \ 2) \ + \ \frac{4}{9}X \ - \ 2 }{X \ + \ ( \frac{4}{9}X \ - \ \frac{14}{3}) } \ = \ - \frac{s_1}{s_2} $$
$$ \Rightarrow \ \ \left( \ \frac{\frac{13}{9}X }{ \frac{13}{9}X \ - \ \frac{14}{3} } \ \right)^2 \ = \ \frac{(X \ + \ 2)^2 + (\frac{2}{3}Y \ - \ 3)^2}{X^2 \ + \ ( \ \frac{2}{3}X \ - \ 7 \ )^2} \ \ . $$
With a bit of help from a computer-algebra system (WA), we obtain two real solutions, $ \ X \ = \ \frac{21}{17} \ $ and the "spurious" solution $ \ -\frac{21}{4} \ $ . The differentiation itself isn't particularly hard; it's the algebra that requires a lot of patience...