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

$\text{Let AD be the bisector of}\angle AThenBD:DC=AB:AC$

$\text{Now,}|AB|=\sqrt{(4-0{)}^2+(3-0{)}^2}=5|AC|=\sqrt{(4-2{)}^2+(3-3{)}^2}=2\Rightarrow \frac{AB}{AC}=\frac{BD}{DC}=\frac{5}{2}\Rightarrow \text{D divides BC in the ratio}5:2\text{So, coordinates of D are}(\frac{5\times 2+0}{5+2},\frac{5\times 3+0}{5+2})=(\frac{10}{7},\frac{15}{7})\therefore \text{The equation of AD is}y-3=\frac{3-\frac{15}{7}}{3-\frac{10}{7}}(x-4)y-3=\frac{21-5}{28-10}(x-4)\Rightarrow y-3=\frac{1}{3}(x-4)\Rightarrow 3(y-3)=x-4\Rightarrow x-3y+9-4=0\Rightarrow x-3y+5=0$