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________

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Aditya Gupta · Answer

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 :))

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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________

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________

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