Straight LineHard
Question
Let $A(1,0),B(2, - 1)$ and $C\left( \frac{7}{3},\frac{4}{3} \right)$ be three points. If the equation of the bisector of the angle ABC is $\alpha x + \beta y = 5$, then the value of $\alpha^{2} + \beta^{2}$ is
Options
A.8
B.5
C.13
D.10
Solution
$${\frac{BD}{DC} = \frac{AB}{AC} = \frac{\sqrt{2} \times 3}{5\sqrt{2}} = \frac{3}{5} }{D = \left( \frac{12}{8},\frac{4}{8} \right) = \left( \frac{3}{2},\frac{1}{2} \right) }$$Slope of $AD = \frac{- 3/2}{\frac{1}{2}} = - 3$
$${3x + y = 5 }{\alpha = 3,\beta = 1;\alpha^{2} + \beta^{2} = 10}$$
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