Question:
If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1(c2 - c3) + m2 (c3 - c1) + m3 (c1 - c2) = 0.
Answer:
Given equations are:
y = m1 x + c1
=> m1 x - y + c1 = 0..........1
y = m2 x + c2
=> m2 x - y + c2 = 0..........2
y = m3 x + c3
=> m3 x - y + c3 = 0..........3
Now three lines are concurrent if
|m1 -1 c1 |
|m2 -2 c2 | = 0
|m3 -1 c3 |
=> m1 (-c3 + c2 ) - m2 (-c3 + c1 ) + m3 (-c2 + c1 ) = 0
=> m1 (c2 - c3 ) + m2 (c3 - c1 ) + m3 (c1 - c2 ) = 0
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