Q.8 A triangle is formed by the lines whose equations are AB: x + y -5 =0, BC : x + 7y-7 = 0 and CA: 7x + y-14 = 0. Find the bisector of the interior angle at B and the exterior angle at C. Determine the nature of the interior angle at A and find the equation of the bisector.

The equation of bisector of angle B will bex+y-51+1=±x+7y-71+49Now taking +ve sign for calculating interior angle bisector.x+y-52=+x+7y-7525x+5y-25=x+7y-74x-2y-18=0Also for exterior angle bisector of Cx+7y-71+49=-7x+y-141+49 -ve sign will be taken for exterior bisectorx+7y-7=-7x-y+148x+8y-21=0 Similarly you may find bisector of angle A. Regards

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