Answer to Question #112822 in Analytic Geometry for SAmuel

Question

A triangle has vertices A(-16,0) B(9,0) c(0,12).prove that the equation of the internal bisector of angle BAC of the triangle is x-3y+16=0

Answer 1

Angle A is formed by straight lines AB and AC. We compose the equations of these lines.

(y-y₁)/(y₂-y₁) = (x-x₁)/(x₂-x₁)

The equation of the line AB: y*(9+16)=(x+16)*0; 25y=0; y=0

The equation of the line AC: y/12=(x+16)/16; 16y=12*x+192; 3x-4y+48=0

Angle bisector equation: (a₁*x+b₁*y+c₁)/ $\sqrt{a_1^2+b_1^2}$ =$\pm$ (a₂*x+b₂*y+c₂)/ $\sqrt{a_2^2+b_2^2}$

We substitute the equations of lines AB and AC into the formulas of the equation of the angle bisectors:

y/1=$\pm$ (3x-4y+48)/$\sqrt{25}$ )y=$\pm$ (3x-4y+48)

So, 3x-9y+48=0 and 3x+y-48=0

Of these equations, the BAC is the equation of the bisector of the internal angle of the triangle BAC, the other is the bisector of the external angle at the vertex A. How to distinguish the equation of the bisector of the internal angle?

Points B and C lie on one side of the bisector of the external angle, therefore, when substituting the coordinates B and C in the equation, we get numbers of the same sign. The bisectors of the inner angle B and C lie on opposite sides, therefore, substituting their coordinates in the equation of the bisector of the inner angle gives us numbers of different signs.

In this way, 3x-9y+48=0 bisector equation of the internal angle BAC