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Line x/a+y/b=1 cuts the co-ordinate axes...

Line `x/a+y/b=1` cuts the co-ordinate axes at A(a,0) and B(0,b) and the line `x/a'+y/b'=-1` at `A'(-a',0)` and `B'(0,-b')`. If the points A,B,A',B' are concyclic then the orthocentre of triangle ABA' is

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Line x/a+y/b=1 cuts the coordinate axes at A(a,0) and B(0,0) and the line x/a+y/b= -1 at A'(-a',0) and B'(0,-b') . If the points A,B A' ,B' are concyclic , then the orthocentre of the triangle ABA' is

Line x/a+y/b=1 cuts the coordinate axes at A(a ,0)a n dB(0,b) and the line x/a^(prime)+y/b^(prime)=-1 at A (-a ,) and B^(prime)(0,-b^(prime))dot If the points A ,B ,A^(prime),B ' are concyclic, then the orthocentre of triangle A B A ' is (0,0) (b) (0,b^(prime)) (0,(a a^(prime))/b) (d) (0,(b b^(prime))/a)

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The algebraic sum of distances of the line ax + by + 2 = 0 from (1,2), (2,1) and (3,5) is zero and the lines bx - ay + 4 = 0 and 3x + 4y + 5=0 cut the coordinate axes at concyclic points. Then (a) a+b=-2/7 (b) area of triangle formed by the line ax+by+2=0 with coordinate axes is 14/5 (c) line ax+by+3=0 always passes through the point (-1,1) (d) max {a,b}=5/7

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Statement 1: If the lines 2x+3y+19=0 and 9x+6y-17=0 cut the x -axis at A,B and the y-axis at C,D, then the points,A,B,C,D are concyclic.Statement 2: since OAxOB=OCxOD, where O is the origin,A,B,C,D are concyclic.