coordinate geometry

Feb 2010
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the equations of side ab ac and bc are 3x+4y=6 ,12x-5y=3 and 4x-3y+12=0 find the equation of internal angle bisector of a

i have solved this through incentre method but it is getting too lengthy.can yu tell me some shorter method may be angle bisector method.
 

earboth

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the equations of side ab ac and bc are 3x+4y=6 ,12x-5y=3 and 4x-3y+12=0 find the equation of internal angle bisector of a

i have solved this through incentre method but it is getting too lengthy.can yu tell me some shorter method may be angle bisector method.
If - and only if - you are allowed to use vectors:

1. The angle at A is produced by the lines AB and AC, which have the normal vectors (3, 4) and (12, -5) respectively.

2. The angle between the 2 normal vectors equals the angle between the 2 lines at the vertex A. Use the Cosine rule to determine the angle:

3. \(\displaystyle \cos(\angle( A)) = \dfrac{(3,4) \cdot (12,-5)}{|(3,4)| \cdot |(12,-5)|} = \dfrac{16}{65}~\implies~|\angle (A)| \approx 75.74997^\circ\)
 
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