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.

$$\begin{gathered} AB=\sqrt{{\left(4-0\right)}^2+{\left(3-0\right)}^2}=5 \\ AC=\sqrt{{\left(4-2\right)}^2+{\left(3-3\right)}^2}=2 \end{gathered}$$

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

$$\begin{gathered} y-3=\frac{3-{\displaystyle \frac{15}{7}}}{4-{\displaystyle \frac{10}{7}}}\left(x-4\right) \\ \Rightarrow y-3=\frac{1}{3}\left(x-4\right) \\ \Rightarrow x-3y+5=0 \end{gathered}$$