Vertices A(1,1),B(4,-2) and C(5,5) of a triangle are given.can you find the equation of the perpendicular dropped from C to the internal bisector of angle A?

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Answer 1 · 2018-03-11T10:02:36

See below.

With #p = (x,y)#

the bisector line is given by

#L_1->p = A+lambda vec v#

where

#vec v = (B-A)/norm(B-A) + (C-A)/norm(C-A) = (sqrt2,0)#

The perp line is given by

#L_2->p = C+lambda vec w#

where

#vec w = (v_y,-v_x) = (0,-sqrt2)#

because then

#<< vec v, vec w >> = 0#

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