Question
If three lines whose equations are $y=m_1 x+c_1$ $y=m_2 x+c_2$ and $y=m_3 x+c_3$ are concurrent, then show that $m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right)=0$.

Answer

The equation of the given lines are
$y=m_1 x+c_1\ldots(i)$
$\begin{array}{l}
y=m_2 x+c_2\ldots(ii) \\
y=m_3 x+c_3\ldots(iii)
\end{array}$
On subtracting equation (i) from (ii), we obtain
0 = (m2 - m1 x + (c2 - c1)
⇒(m1 - m2)x = c2 - c1
⇒$x=\frac{c_2-c_1}{m_1-m_2}$
On substituting this value of x in (i), we obtain
$y=m_1\left(\frac{c_2-c_1}{m_1-m_2}\right)+c_1$
$y=\frac{m_1 c_2-m_1 c_1}{m_1-m_2}+c_1$
$y=\frac{m_1 c_2-m_1 c_1+m_1 c_1-m_2 c_1}{m_1-m_2}$
$y=\frac{m_1 c_2-m_2 c_1}{m_1-m_2}$
$\therefore \quad\left(\frac{c_2-c_1}{m_1-m_2}, \frac{m_1 c_2-m_2 c_1}{m_1-m_2}\right)$ is the point of intersection of lines (i) and (ii).
It is given that lines (i), (ii) and (iii) are concurrent. Hence, the point of intersection of lines (i) and (ii) will also satisfy equation (iii),
$\frac{m_1 c_2-m_2 c_1}{m_1-m_2}=m_3\left(\frac{c_2-c_1}{m_1-m_2},\right)+c_3$
$\frac{m_1 c_2-m_2 c_1}{m_1-m_2}=\frac{m_3 c_2-m_3 c_1+c_3 m_1-c_3 m_2}{m_1-m_2}$
$\begin{aligned} m_1 c_2-m_2 c_1-m_3 c_2+m_3 c_1-c_3 m_1+c_3 m_2 & =0 \\ m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right) & =0\end{aligned}$
Hence, $m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right)=0$.

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