Question

If A(4, 3), B(0, 0), and C(2, 3) are the vertices of ∆ABC then find the equation of bisector of angle BAC.

Answer 1

A(4, 3), B(0, 0) and C(2, 3) are the vertices of ΔABC.

Let AD be the bisector of angle BAC.

Then D divides side BC in the ratio AB : AC.

Now, AB = ${\displaystyle \sqrt{{(4-0)}^2+{(3-0)}^2}}$

= ${\displaystyle \sqrt{16+9}}$

= ${\displaystyle \sqrt{25}}$

= 5

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

= ${\displaystyle \sqrt{4+0}}$

= ${\displaystyle \sqrt{4}}$

= 2

∴ D divides BC internally in the ratio 5 : 2, where B(0, 0) and C(2, 3).

∴ by section formula,

D ≡ ${\displaystyle (\frac{5\times 2+2\times 0}{5+2},\frac{5\times 3+2\times 0}{5+2})=(\frac{10}{7},\frac{15}{7})}$

∴ equation of the angle bisector AD is

${\displaystyle \frac{y-3}{x-4}=\frac{\frac{15}{7}-3}{\frac{10}{7}-4}}$

∴ ${\displaystyle \frac{y-3}{x-4}=\frac{15-21}{10-28}=\frac{-6}{-18}=\frac{1}{3}}$

∴ 3y – 9 = x – 4

∴ x – 3y + 5 = 0.