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```        A(1,3) and C(-2/5,-2/5) are vertices of triangle ABC and the equation of internal angle bisector of  angle ABC is x+y=2 then
1)equation of side BC is
a)7x+3y-4=0
b)7x+3y+4=0
c)7x-3y+4=0
d)7x-3y-4=0

2)coordinates of vertex B is
a)(3/10,7/10)
b(17/10,3/10)
c)(-5/2,9/2)
d)(1,1)

3)equation of side AB is
a)3x+7y=24
b)3x+7y=-24
c)13x+7y+8=0
d)13x-7y+8=0
```
8 years ago

```							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.
```
8 years ago
```							draw a line A’on the side bc .from this ucan find the image of A which lies on bc use the formula and get the points then use two point form and ur equation is ready
```
one year ago
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