Brilli's faithfulness for Billi

The shortest distance $B_1P + PB_2$ follows the principle of least action. It is that traveled by light from point $B_1$ to point $P$ and then reflected at point $P$ (as if the straight-line ( $2x+y-6=0$ ) river is a mirror) to point $B_2$ .

Let the points where altitudes from $B_1$ and $B_2$ meet $2x+y-6=0$ be $N_1$ and $N_2$ respectively. Since the angle of incident and angle of reflection is the same, $\triangle B_1PN_1$ and $\triangle B_2PN_2$ are similar and hence, $\frac{PN_1}{PN_2} = \frac{B_1N_1}{B_2N_2}$ .

Since the river $2x+y-6=0 \Rightarrow y = -2x+6$ has a gradient of $-2$ , lines perpendicular to it must have a gradient of $\frac{1}{2}$ . Therefore the equation of $B_1N_1$ is given by:

$\dfrac {y-10}{x-8} = \dfrac {1}{2} \quad \Rightarrow y = \frac{1}{2}x + 6$

The coordinates of $N_1$ is given by:

$y = -2x+6 = \frac{1}{2}x + 6 \quad \Rightarrow x_1 = 0$ and $y_1 = 6$ .

The length of $B_1N_1 = \sqrt{(8-0)^2+(10-6)^2} = \sqrt{80}$

Similarly, $B_2N_2: \dfrac {y-30}{x+2} = \dfrac {1}{2} \quad \Rightarrow y = \frac{1}{2}x + 31$

The coordinates of $N_2: y = -2x+6 = \frac{1}{2}x + 31 \quad \Rightarrow x_2 = -10$ and $y_2 = 26$ .

The length of $B_2N_2 = \sqrt{(-2+10)^2+(30-26)^2} = \sqrt{80}$

Since $B_1N_1 = B_2N_2\quad \Rightarrow PN_1 = PN_2$ and $P$ is midway between $N_1$ and $N_2$ and its coordinates:

$x_P = \dfrac {x_1+x_2}{2} = \dfrac{0-10}{2} = -5 \quad y_P = \dfrac {y_1+y_2}{2} = \dfrac{6+26}{2} = 16$

$\Rightarrow x_P + y_P =-5+26 = \boxed{21}$

What's the correct answer? 11 or 21?