JEE Mains · Maths · STD 11 - 9. straight line
Let \( A(1,0) \), \( B(2,-1) \) and \( C(\frac{7}{3},\frac{4}{3}) \) 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
- A 8
- B 5
- C 13
- D 10
Answer & Solution
Correct Answer
(D) 10
Step-by-step Solution
Detailed explanation
\(\frac{ BD }{ DC }=\frac{ AB }{ AC }=\frac{\sqrt{2} \times 3}{5 \sqrt{2}}=\frac{3}{5}\) \(D =\left(\frac{12}{8}, \frac{4}{8}\right)=\left(\frac{3}{2}, \frac{1}{2}\right)\) Slope of \(AD =\frac{-3 / 2}{\frac{1}{2}}=-3\) \(3 x+y=5\) \(\alpha=3, \beta=1 ; \alpha^2+\beta^2=10\)
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