The given equation of lines are y = $m_1 x + c_1$ ... (i) y = $m_2 x + c_2$ ... (ii) y = $m_3 x + c_3$ ... (iii) On solving (i) and (ii), we get x = $\frac{-(c_2-c_1)}{m_2-m_1}$ , y = $\frac{c_1 m_2 - c_2 m_1}{m_2 - m_1}$ Hence, (i) & (ii) meets at $\left[ \frac{-(c_1-c_2)}{m_1-m_2} , \frac{m_1 c_2 - m_2 c_1}{m_1-m_2} \right]$ Clearly (i), (ii) and (iii) lines are concurrent if the above point lies on (iii) i.e., if $\left[ \frac{m_1 c_2 - m_2 c_1}{m_1-m_2} \right]$ = $m_2 \left[ \frac{-(c_1-c_2)}{m_1-m_2} \right] + c_3$ i.e., if $\frac{m_1 c_2 - m_2 c_1}{m_1-m_2}$ = $\frac{m_3 (-(c_1-c_2))}{m_1-m_2} + c_3$ i.e., if $(m_1 c_2 - m_2 c_1)$ = $m_3 (c_2 - c_1) + c_3 (m_1 - m_2)$ ⇒ $m_1 c_2 - m_2 c_1$ = $m_3 c_2 - m_3 c_1 + c_3 m_1 - c_3 m_2$ ⇒ $m_1 c_2 - m_2 c_1 - m_3 c_2 + m_3 c_1 - c_3 m_1 + c_3 m_2$ = 0 ⇒ $m_1 (c_2 - c_3) + m_2 (c_3 - c_1) + m_3 (c_1 - c_2)$ = 0. Hence proved.