Given lies are

$L_1: y=m x+a$

$L_2: y=n x+b$

Line $L_1$ intersects $y-$ axis at $A=(0, a)$

Line $L_2$ intersects $y-$ axis at $B=(0, b)$

Comparing the given two equations,

$m x+a=n x+b$

$\Rightarrow n x-m x=a-b$

$\Rightarrow x=\frac{a-b}{n-m}$

Substituting above value in $L_1$ we get,

$y=m \frac{a-b}{n-m}+a$

$\Rightarrow y=\frac{m a-m b+n a-m a}{n-m}$

$\Rightarrow y=\frac{n a-m b}{n-m}$

$\therefore$ The co-ordinates of intersection of the lines is $C =\left(\frac{a-b}{ n – m }, \frac{ na – mb }{ n – m }\right)$

Distance between $A$ and $B$

$d=\sqrt{(0-b)^2+(a-b)^2}$

$\Rightarrow d=\sqrt{(a-b)^2}$

$\Rightarrow d=a-b$

Length of perpendicular from $C$ on the $y$ – axis.

$p=\frac{a-b}{n-m}$

$\therefore$ Area of the triangle fromed by the lines $L _1, L _2$ and $y -$ axis

$A=\left|\frac{1}{2} d p\right|$

$\Rightarrow A=\frac{1}{2} \times(a-b) \times \frac{a-b}{n-m}$