Question

Find the equation of internal bisector of angle bac of triangle abc whose vertices are a (5,2) b(2,3) c (6,5)

Answer 1

Answer:    

Step-by-step explanation: Let the given vertices be A(5,2), B(2,3) and C(6,5).

Equation of AB is given by

$y-2=-\dfrac{1}{3}(x-5)\\\Rightarrow 3y-6=-x+5\\\Rightarrow x+3y-11=0$

and equation of AC is

$y-2=3(x-5)\\\Rightarrow3x-y-13=0.$

Now, if (x,y) is any point on the bisector AD of angle A, then

$| \dfrac{x+3y-11}{\sqrt{1^{2} +3^{2} } } |=3x-y-13,$which gives the equation of internal bisector if we assume positive value of the modulus.

Thus, after solving, we will get

$x+3y-11=3\sqrt{10} x-\sqrt{10}y-13\sqrt{10}\\\Rightarrow (3\sqrt{10}-1 )x-(3+\sqrt{10} )y-13\sqrt{10}+11=0,$ which is the required equation of the bisector.