If three lines whose equations are

Solution:

The equations of the given lines are

y = m1x + c1 … (1)

y = m2x + c2 … (2)

y = m3x + c3 … (3)

On subtracting equation (1) from (2), we obtain

$0=\left(m_{2}-m_{1}\right) x+\left(c_{2}-c_{1}\right)$\

$\Rightarrow\left(m_{1}-m_{2}\right) x=c_{2}-c_{1}$

$\Rightarrow x=\frac{c_{2}-c_{1}}{m_{1}-m_{2}}$

On substituting this value of x in (1), we obtain

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

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

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

$y=\frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}$

$\therefore\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}, \frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}\right)$ is the point of intersection of lines $(1)$ and $(2)$.

It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of lines (1) and (2) will also satisfy equation (3).

$\frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}=m_{3}\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\right)+c_{3}$

$\frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}=\frac{m_{3} c_{2}-m_{3} c_{1}+c_{3} m_{1}-c_{3} m_{2}}{m_{1}-m_{2}}$

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

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

Hence, $m_{1}\left(c_{2}-c_{3}\right)+m_{2}\left(c_{3}-c_{1}\right)+m_{3}\left(c_{1}-c_{2}\right)=0$