Question

The equation of the internal bisector of $\angle$BAC of $\Delta$ABC with vertices A(5, 2), B(2, 3) and C(6, 5), is

• A. 2x + y + 12 = 0
• B. x + 2y – 12 = 0
• C. 2x + y – 12 = 0
• D. x + 2y +12 = 0

Answer 1

The correct option is C 2x + y – 12 = 0

Let AD be the internal bisector of angle BAC cutting BC at D.

$Now,AB=\sqrt{(5-2{)}^2+(2-3{)}^2}=\sqrt{10}andAC=\sqrt{(5-6{)}^2+(2-5{)}^2}=\sqrt{10}$

since AD is the internal bisector of angle BAC,

$\therefore \frac{BD}{DC}=\frac{AB}{AC}=\frac{\sqrt{10}}{\sqrt{10}}=\frac{1}{1}$

$\therefore$ Coordinates of D are $(\frac{2+6}{2},\frac{3+5}{2})$ i.e. (4, 4)

So, the equation of AD is

$y-2=\frac{2-4}{5-4}$ (x – 5) or 2x + y – 12 = 0