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?

Ans : 4x + y = 7

1 Answer
Jan 10, 2018

(9x-8y+17)/sqrt29=+-(3x+16y-13)/sqrt539x8y+1729=±3x+16y1353.

Explanation:

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

Then, from Geometry, we know that, PP is equidistant

from the sides AB and ACABandAC.

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

AB : y-1={(1-(-8))/(-1-(-9))}(x-(-1)), i.e., AB:y1={1(8)1(9)}(x(1)),i.e.,

AB : 8(y-1)=9(x+1), or, 9x-8y+17=0AB:8(y1)=9(x+1),or,9x8y+17=0.

Similarly, AC : 3x+16y-13=0AC:3x+16y13=0.

Now, the bot-"distance "d_1 distance d1 from P" to "ABP to AB is given by,

d_1=|9x-8y+17|/sqrt(9^2+(-8)^2)=|9x-8y+17|/sqrt145d1=|9x8y+17|92+(8)2=|9x8y+17|145.

The bot-"distance "d_2 distance d2 from P" to "ACP to AC is,

d_2=|3x+16y-13|/sqrt(3^2+16^2)=|3x+16y-13|/sqrt265d2=|3x+16y13|32+162=|3x+16y13|265.

But, d_1=d_2d1=d2.

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

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

eqns.