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Dear student,
Let the three given lines be
a1x+b1y+c1=0 ...(i)
a2x+b2y+c2=0 ..(ii)
a3x+b3y+c3=0 ...(iii)
For the given lines to be concurrent, no two of these lines can be parallel or coincident
b1a1=b2a2=b3a3 ...(iv)
and the point of intersection of any two lines must lie on the third. The point of intersection of (i) and (ii) is
b1c2−b2c1a1b2−a2b1c1a2−c2a1a1b2−a2b1 (obtain it)
This point must lie on the third line, so we must have
a3b1c2−b2c1a1b2−a2b1+b3c1a2−c2a1a1b2−a2b1+c3=0
a3b1c2−a3b2c1+a2b3c1−a1b3c2+a1b2c3−a2b1c3=0
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