Question

Find the equation of the bisector of angle A of the triangle whose vertices are A (4, 3), B (0, 0) and C(2, 3).

Answer 1

The vertices of triangle ABC are A (4, 3), B (0, 0) and C (2, 3).

Let us find the lengths of sides AB and AC.

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

\[AC = \sqrt{\left( 4 - 2 \right)^2 + \left( 3 - 3 \right)^2} = 2\]

We know that the internal bisector AD of angle BAC divides BC in the ratio AB : AC  i.e.  5 : 2

\[\therefore D \equiv \left( \frac{2 \times 0 + 5 \times 2}{5 + 2}, \frac{2 \times 0 + 5 \times 3}{5 + 2} \right) = \left( \frac{10}{7}, \frac{15}{7} \right)\]

Thus, the equation of AD is

\[y - 3 = \frac{3 - \frac{15}{7}}{4 - \frac{10}{7}}\left( x - 4 \right)\]

\[ \Rightarrow y - 3 = \frac{1}{3}\left( x - 4 \right)\]

\[ \Rightarrow x - 3y + 5 = 0\]