Three vertices of triangle ABC are A(-1 ,1 ), B (-9,-8) and C (15,-2) then the equation of angle bisector of angle A is?

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Ans : 4x + y = 7

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Answer 1 · 2018-01-10T10:49:17

# (9x-8y+17)/sqrt29=+-(3x+16y-13)/sqrt53#.

Let #P(x,y)# be any point on the #/_-"bisector"# of #/_A#.

Then, from Geometry, we know that, #P# is equidistant

from the sides #AB and AC#.

#A(-1,1), & B(-9,-8) rArr the eqn. of #AB# is given by,

#AB : y-1={(1-(-8))/(-1-(-9))}(x-(-1)), i.e., #

#AB : 8(y-1)=9(x+1), or, 9x-8y+17=0#.

Similarly, #AC : 3x+16y-13=0#.

Now, the #bot-"distance "d_1 # from #P" to "AB# is given by,

#d_1=|9x-8y+17|/sqrt(9^2+(-8)^2)=|9x-8y+17|/sqrt145#.

The #bot-"distance "d_2 # from #P" to "AC# is,

#d_2=|3x+16y-13|/sqrt(3^2+16^2)=|3x+16y-13|/sqrt265#.

But, #d_1=d_2#.

#:. |9x-8y+17|/sqrt145=|3x+16y-13|/sqrt265#.

#:. (9x-8y+17)/sqrt29=+-(3x+16y-13)/sqrt53#, are the desired

eqns.