JEE Mains · Maths · STD 11 - 9. straight line
Let \(\mathrm{ABC}\) be a triangle with \(\mathrm{A}(-3,1)\) and \(\angle \mathrm{ACB}=\theta, 0<\theta<\frac{\pi}{2} .\) If the equation of the median through \(\mathrm{B}\) is \(2 \mathrm{x}+\mathrm{y}-3=0\) and the equation of angle bisector of \(\mathrm{C}\) is \(7 \mathrm{x}-4 \mathrm{y}-1=0\) then \(\tan\, \theta\) is equal to:
- A \(\frac{1}{2}\)
- B \(\frac{3}{4}\)
- C \(\frac{4}{3}\)
- D \(2\)
Answer & Solution
Correct Answer
(C) \(\frac{4}{3}\)
Step-by-step Solution
Detailed explanation
\(\therefore \quad \mathrm{M}\left(\frac{\mathrm{a}-3}{2}, \frac{\mathrm{b}+1}{2}\right)\) lies on \(2 \mathrm{x}+\mathrm{y}-3=0\) \(\Rightarrow 2 \mathrm{a}+\mathrm{b}=11 \ldots \ldots \ldots . . (i)\) \(\because \mathrm{C}\) lies on \(7 \mathrm{x}-4 \mathrm{y}=1\)…
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