Show that the area of the triangle formed by the lines

$y=m_{1} x+c_{1}, y=m_{2} x+c_{2}$ and $x=0$ is $\frac{\left(c_{1}-c_{2}\right)^{2}}{2\left|m_{1}-m_{2}\right|}$.

Given lines are

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

$y=m_{1} x+c_{2}$.....$(2)$

$x=0$.....$(3)$

We know that line $y=m x+c$ meets the line $x=0$ ($y-$ axis) at the point $(0, c) .$ Therefore, two vertices of the triangle formed by lines $(1)$ to $(3)$ are $\left. P \left(0, c_{1}\right) \text { and } Q \left(0, c_{2}\right) \text { (Fig } .\right)$

Third vertex can be obtained by solving equations $( 1 )$ and $( 2 )$. Solving $(1)$ and $(2)$, we get

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

Now, the area of the triangle is

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