let ABC be a triangle such that the coordinates of vertex A are (-3,1).equation of the median of B is 2x+y-3=0 and equation of the angular bisector of C is 7x-4y-1=0. Find the slope of the line BC

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Let co-ordinates  of C be r,s.Co-ordinates of A are A-3,1Equation of median of B is 2x+y-3=0 and equation of angular bisector of C is 7x-4y-1=0.Point C will lie on the angular bisector of C.7r-4s-1=07r-4s=1 ;equationiMid point of AC will be r-32,s+12. Now this mid point will lie on median through B.2r-32+s+12-3=02r-6+s+1-6=02r+s=11equationi+4×equationii, we get,7r-4s+8r+4s=1+4415r=45r=3Putting this value in equationii, we get,2×3+s=11s=5r,s=3,5Slope of AC m1=5-13--3=46=23Slope of bisector 7x-4y-1=0 m2 is -Cofficient of xCoefficient of y=-7-4=74Angle between AC and bisector through C will be:tanα=m1-m21+m1.m2=23-741+23.74=8-211212+1412=-1326tanα=12Let slope of BC be m.Angle between BC and bisector through C=Angle between AC and bisector through Cm-m21+m.m2=12m-741+m.74=124m-77m+4=124m-77m+4=±124m-77m+4=-12 or 4m-77m+4=128m-14=-7m-4 or 8m-14=7m+48m+7m=14-4 or 8m-7m=14+415m=10 or m=18m=23 or m=18 Hence slope of BC will be either be 23 or 18.But as slope of AC is also 23 so when BC=23, A,B,C will become collinear and no triangle will be formed.Hence slope of BC for given case will be 18. 

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