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 = `sqrt((4 - 0)^2 + (3 - 0)^2)`

= `sqrt(16 + 9)`

= `sqrt(25)`

= 5

and AC = `sqrt((4 - 2)^2 + (3 - 3)^2`

= `sqrt(4 + 0)`

= `sqrt(4)`

= 2

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

∴ by section formula,

D ≡ `((5 xx 2 + 2 xx 0)/(5 + 2), (5 xx 3 + 2 xx 0)/(5 + 2)) = (10/7, 15/7)`

∴ equation of the angle bisector AD is

`(y - 3)/(x - 4) = (15/7 - 3)/(10/7 - 4)`

∴ `(y - 3)/(x - 4) = (15 - 21)/(10 - 28) = (-6)/(-18) = 1/3`

∴ 3y – 9 = x – 4

∴ x – 3y + 5 = 0.