ExamBro
ExamBro
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:

  1. A \(\frac{1}{2}\)
  2. B \(\frac{3}{4}\)
  3. C \(\frac{4}{3}\)
  4. D \(2\)
Verified Solution

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 \(7 \mathrm{x}-4 \mathrm{y}=1\)…