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# The equation of sides of a triangle ABC are AB: x+y-1=0 , BC: 7x-y-15=0, AC : x-y-1=0. The equation of the angular bisector of interval angle B of the triangle is 3x+y-k =0. The positive integer k is________

2075 Points
one year ago
the property of angle bisector of 2 lines L1 and L2 that reflection of any point on L1 about the bisector lies on L2 shall be used here.
AB: x+y-1=0 , BC: 7x-y-15=0
consider a pt P(1, 0) on AB. let its reflection about La: 3x+y-k =0 be R(2p – 1, 2q).
then mid pt M of P and R would lie on La. clearly by section formula M is (p, q).
so 3p+q= k...(1)
also, PR will be perpendicular to La.
so ( – 3)*(2q/(2p-1 – 1))= – 1
or 3q= p – 1.......(2)
solve (1) and (2), we get
p= (3k+1)/10, q= (k – 3)/10
so, R(2p – 1, 2q)= ((3k – 4)/5, (k – 3)/5).
since R should lie on BC, we have
7(3k – 4)/5= (k – 3)/5 + 15
or k= 5
KINDLY APPROVE :))