Question

Let \(ABC\) be a triangle with \(A(-3, 1)\) and \(\angle ACB = \theta, 0<\theta \leq \frac{\pi}{2}\). If the equation of the median through \(B\) is \(2x + y - 3 = 0\) and the equation of angle bisector of \(C\) is \(7x - 4y - 1 = 0\), then \(\tan \theta\) is equal to :

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

Hint:

For problems involving vertex angles and bisectors, finding the angle between the side and the bisector (\(\theta/2\)) is often the most direct path to the solution.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We use the properties of medians and angle bisectors. Let the coordinates of vertex \(C\) lie on the angle bisector. The midpoint of \(AC\) lies on the median from \(B\). Using these constraints, we find the coordinates of \(C\) and then the angle \(\theta\).
Step 2: Key Formula or Approach:
1. Coordinates of \(C\) on \(7x - 4y - 1 = 0\): \(C(t, \frac{7t-1}{4})\).
2. Midpoint \(M\) of \(AC\): \(\left(\frac{t-3}{2}, \frac{7t+3}{8}\right)\).
3. \(M\) lies on \(2x + y - 3 = 0\).
4. Angle between two lines with slopes \(m_1\) and \(m_2\): \(\tan \phi = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|\).
Step 3: Detailed Explanation: Substitute \(M\) into the median equation: \[ 2\left(\frac{t-3}{2}\right) + \frac{7t+3}{8} - 3 = 0 \implies t-3 + \frac{7t+3}{8} - 3 = 0 \] \[ 8t - 48 + 7t + 3 = 0 \implies 15t = 45 \implies t = 3 \] So, \(C(3, 5)\).
Slope of \(AC\), \(m_{AC} = \frac{5-1}{3 - (-3)} = \frac{4}{6} = \frac{2}{3}\).
Slope of the angle bisector of \(C\), \(m_L = \frac{7}{4}\).
The angle between \(AC\) and the bisector is \(\theta/2\). \[ \tan(\theta/2) = \left| \frac{7/4 - 2/3}{1 + (7/4)(2/3)} \right| = \left| \frac{13/12}{(12+14)/12} \right| = \frac{13}{26} = \frac{1}{2} \] Now, find \(\tan \theta\): \[ \tan \theta = \frac{2\tan(\theta/2)}{1 - \tan^2(\theta/2)} = \frac{2(1/2)}{1 - (1/4)} = \frac{1}{3/4} = \frac{4}{3} \]
Step 4: Final Answer:
The value of \(\tan \theta\) is \(\frac{4}{3}\).