A(1,3) and C(-2/5,-2/5) are vertices of triangle ABC and the equation of internal angle bisector ofangle ABC is x+y=2 then1)equation of side BC isa)7x+3y-4=0b)7x+3y+4=0c)7x-3y+4=0d)7x-3y-4=02)coordinates of vertex B isa)(3/10,7/10)b(17/10,3/10)c)(-5/2,9/2)d)(1,1)3)equation of side AB isa)3x+7y=24b)3x+7y=-24c)13x+7y+8=0d)13x-7y+8=0

Dear student,

If the equations of 2 sides of the triangle (say, AB and AC) are
Line1: A1x+B1y+C1=0 and
Line2: A2x+B2y+C2=0
Then (at vertex A)
Angle bisector1: f1(x,y)=(A1x+B1y+C1)/√(A1²+B1²) + (A2x+B2y+C2)/√(A2²+B2²)=0 ... (1)
Angle bisector2: f2(x,y)=(A1x+B1y+C1)/√(A1²+B1²) - (A2x+B2y+C2)/√(A2²+B2²)=0 ... (2)

To find out which one is the internal angle bisector at, say, vertex A, you'll have to substitute the coordinates of the other 2 vertices, B and C, in one of the equations (1) or (2). The internal angle bisector corresponds to the equation, which will give
f(B)*f(C)<0. For the external angle bisector it holds f(B)*f(C)>0.

Approved