Question

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

Answer 1

AD is the bisector of $\angle A$

$\therefore \frac{AB}{AC}=\frac{BD}{CD}$ [Internal angle bisector theorem]

$AB=\sqrt{(4-0{)}^2+(3-0{)}^2}=\sqrt{16+9}=\sqrt{25}=5$

$AC=\sqrt{(4-2{)}^2+(3-3{)}^2}=\sqrt{4+0}=2$

So, $\frac{BD}{CD}=\frac{5}{2}$

$\therefore$ Coordinates of D

$=(\frac{5\times 2+2\times 0}{5+2},\frac{5\times 3+2\times 0}{5+2})$ [Section formula]

$=(\frac{10}{7},\frac{15}{7})$

Equation of the straight line passing through $(x_1,y_1)and(x_2,y_2)is(y-y_1)=\left(\frac{y_2-y_1}{x_2-x_1}\right)(x-x_1)$

$\therefore$ Equation of AD is

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

$(y-3)=\frac{-6}{-18}(x-4)$

$(y-3)=\frac{1}{3}(x-4)$

$3y-9=x-4$

$x-3y+5=0$