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 $\frac{(c_{1}-c_{2})^{2}}{2|m_{1}-m_{2}|}$.

(N/A) Given lines are:
$y=m_{1}x+c_{1}$ ..... $(1)$
$y=m_{2}x+c_{2}$ ..... $(2)$
$x=0$ ..... $(3)$
We know that the line $y=mx+c$ meets the line $x=0$ ($y$-axis) at the point $(0, c)$. Therefore,two vertices of the triangle formed by lines $(1)$,$(2)$,and $(3)$ are $P(0, c_{1})$ and $Q(0, c_{2})$.
The third vertex $R$ is obtained by solving equations $(1)$ and $(2)$:
$m_{1}x+c_{1} = m_{2}x+c_{2}$
$x(m_{1}-m_{2}) = c_{2}-c_{1}$
$x = \frac{c_{2}-c_{1}}{m_{1}-m_{2}}$
Substituting $x$ in $(1)$:
$y = m_{1}\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\right) + c_{1} = \frac{m_{1}c_{2}-m_{1}c_{1}+m_{1}c_{1}-m_{2}c_{1}}{m_{1}-m_{2}} = \frac{m_{1}c_{2}-m_{2}c_{1}}{m_{1}-m_{2}}$
So,$R = \left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}, \frac{m_{1}c_{2}-m_{2}c_{1}}{m_{1}-m_{2}}\right)$.
The area of the triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ is $\frac{1}{2}|x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2})|$.
Here,vertices are $(0, c_{1}), (0, c_{2}), \left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}, \frac{m_{1}c_{2}-m_{2}c_{1}}{m_{1}-m_{2}}\right)$.
Area $= \frac{1}{2} |0(c_{2} - y_{R}) + 0(y_{R} - c_{1}) + \frac{c_{2}-c_{1}}{m_{1}-m_{2}}(c_{1}-c_{2})|$
Area $= \frac{1}{2} |\frac{-(c_{2}-c_{1})^{2}}{m_{1}-m_{2}}| = \frac{(c_{1}-c_{2})^{2}}{2|m_{1}-m_{2}|}$.