Question

A(3,2,0), B(5,3,2), C(-9,6,-3) are three points forming a triangle. AD, the bisector of angle $BAC$ meets BC in D. Find the coordinates of D:

• A. $\left( \frac{19}{8}, \frac{57}{15}, \frac{57}{15} \right)$
• B. $\left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right)$
• C. (2,3,0)
• D. (4,5,6)

Hint:

The angle bisector theorem is a powerful tool for solving triangle-related coordinate geometry problems.

Answer 1

The Correct Option is B

Since \( AD \) is the bisector of \( \angle BAC \), we use the angle bisector theorem:
\[ \frac{BD}{DC} = \frac{AB}{AC} \quad {...(i)} \] Now, we calculate the lengths of \( AB \) and \( AC \).

Calculating \( AB \)

\[ AB = \sqrt{(5 - 3)^2 + (3 - 2)^2 + (2 - 0)^2} \] \[ = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

Calculating \( AC \)

\[ AC = \sqrt{(-9 - 3)^2 + (6 - 2)^2 + (-3 - 0)^2} \] \[ = \sqrt{144 + 16 + 9} \] \[ = \sqrt{169} = 13 \]

Finding the Ratio of \( BD : DC \)

From equation (i): \[ \frac{BD}{DC} = \frac{3}{13} \] Thus, \( D \) divides \( BC \) in the ratio \( 3:13 \).

Finding the Coordinates of \( D \)

Using the section formula: \[ D \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right) \] where \( B(-9,6,-3) \) and \( C(5,3,2) \), and the ratio \( m:n = 3:13 \), \[ x = \frac{3(-9) + 13(5)}{3+13}, \quad y = \frac{3(6) + 13(3)}{3+13}, \quad z = \frac{3(-3) + 13(2)}{3+13} \] \[ = \left( \frac{-27 + 65}{16}, \frac{18 + 39}{16}, \frac{-9 + 26}{16} \right) \] \[ = \left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right) \] Thus, the coordinates of \( D \) are: \[ D \left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right) \]