Let A(1,0), B(2,-1), and C(frac{7}{3},frac{4} | vedclass

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(D) First, calculate the lengths of sides $AB$ and $BC$:$AB = \sqrt{(2-1)^2 + (-1-0)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$$BC = \sqrt{(\frac{7}{3}-2)^2

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(D) First, calculate the lengths of sides $AB$ and $BC$:$AB = \sqrt{(2-1)^2 + (-1-0)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$$BC = \sqrt{(\frac{7}{3}-2)^2 + (\frac{4}{3}-(-1))^2} = \sqrt{(\frac{1}{3})^2 + (\frac{7}{3})^2} = \sqrt{\frac{1}{9} + \frac{49}{9}} = \sqrt{\frac{50}{9}} = \frac{5\sqrt{2}}{3}$By the Angle Bisector Theorem, the bisector of $\angle ABC$ divides the opposite side $AC$ in the ratio $AB:BC = \sqrt{2} : \frac{5\sqrt{2}}{3} = 3:5$.Let $D$ be the point on $AC$ dividing it in ratio $3:5$. Using the section formula:$D = \left( \frac{3(\frac{7}{3}) + 5(1)}{3+5}, \frac{3(\frac{4}{3}) + 5(0)}{3+5} \right) = \left( \frac{7+5}{8}, \frac{4+0}{8} \right) = (\frac{12}{8}, \frac{4}{8}) = (\frac{3}{2}, \frac{1}{2})$.The angle bisector passes through $B(2,-1)$ and $D(\frac{3}{2}, \frac{1}{2})$.The slope $m = \frac{\frac{1}{2} - (-1)}{\frac{3}{2} - 2} = \frac{\frac{3}{2}}{-\frac{1}{2}} = -3$.The equation of the line is $y - (-1) = -3(x - 2) \implies y + 1 = -3x + 6 \implies 3x + y = 5$.Comparing with $\alpha x + \beta y = 5$, we get $\alpha = 3$ and $\beta = 1$.Thus, $\alpha^2 + \beta^2 = 3^2 + 1^2 = 9 + 1 = 10$.

Let $A(1,0)$, $B(2,-1)$, and $C(\frac{7}{3},\frac{4}{3})$ be three points. If the equation of the internal angle bisector of $\angle ABC$ is $\alpha x+\beta y=5$, then the value of $\alpha^2+\beta^2$ is

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