If the three lines $y = m_{1} x + c_{1}, y = m_{2} x + c_{2}$ and $y = m_{3} x + c_{3}$ are concurrent then show that,

$m_{1} (c_{2} - c_{3}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0$

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Solution

The given lines are:

$m_{1} x - y + c_{1} = 0$ .....(i)

$m_{2} x - y + c_{2} = 0$ .....(ii)

$m_{3} x - y + c_{3} = 0$ .....(iii)

On solving (i) and (ii) by cross multiplication, we get,

$\frac{x}{(- c_{2} + c_{1})} = \frac{y}{(m_{2} c_{1} - m_{1} c_{2})} = \frac{1}{(- m_{1} + m_{2})}$

$\Rightarrow x = \frac{(c_{1} - c_{2})}{(m_{2} - m_{1})} and y = \frac{(m_{2} c_{1} - m_{1} c_{2})}{(m_{2} - m_{1})}$

Thus, the point of intersection of(i) and (ii) is,

P $(\frac{c_{1} - c_{2}}{m_{2} - m_{1}}, \frac{m_{2} c_{1} - m_{1} c_{2}}{m_{2} - m_{1}})$

Since the given three lines meet at a point, the point P must therefore lie on (iii) also.

$∴ m_{3} . \frac{(c_{1} - c_{2})}{(m_{2} - m_{1})} - \frac{(m_{2} c_{1} - m_{1} c_{2})}{(m_{2} - m_{1})} + c_{3} = 0$

$\Rightarrow m_{3} (c_{1} - c_{2}) - (m_{2} c_{1} - m_{1} c_{2}) + c_{3} (m_{2} - m_{1}) = 0$

$\Rightarrow m_{1} (c_{2} - c_{3}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0$

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If the three lines 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

If the three lines $y = m_{1} x + c_{1}, y = m_{2} x + c_{2}$ and $y = m_{3} x + c_{3}$ are concurrent then show that,

$m_{1} (c_{2} - c_{3}) + m_{2} (c_{3} - c_{1}) + m_{3} (c_{1} - c_{2}) = 0$