Let A(1,0), B(2,-1) and C ((7)/(3), (4)/(3)) be three points. If...

For angle bisector of angle ABC where A(1,0), B(2,-1), C($\frac{7}{3}$, $\frac{4}{3}$).
Direction BA: $(1-2, 0-(-1)) = (-1, 1)$, unit vector \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$<br>Direction BC: \left(\frac{1}{3}, \frac{7}{3}\right)$, magnitude $\frac{5\sqrt{2}}{3}$, unit vector \left(\frac{1}{5\sqrt{2}}, \frac{7}{5\sqrt{2}}\right)$<br>Angle bisector direction (sum of unit vectors): \left(\frac{-4}{5\sqrt{2}}, \frac{12}{5\sqrt{2}}\right)$ → direction $(-1, 3)$
Normal to bisector: $(3, 1)$. Bisector passes through B(2, -1):
$3(x-2) + 1(y+1) = 0$ → $3x + y - 5 = 0$
Therefore $\alpha = 3$, $\beta = 1$, so $\alpha^2 + \beta^2 = 9 + 1 = 10$