Given lies are
L1:y=mx+a
L2:y=nx+b
Line L1 intersects y−axis at A=(0,a)
Line L2 intersects y−axis at B=(0,b)
Comparing the given two equations,
mx+a=nx+b
⇒ nx−mx=a−b
⇒ x=a−bn−m
Substituting above value in L1 we get,
y=ma−bn−m+a
⇒ y=ma−mb+na−man−m
⇒ y=na−mbn−m
∴ The co-ordinates of intersection of the lines is C=(a−bn−m,na−mbn−m)
Distance between A and B
d=√(0−0)2+(a−b)2
⇒ d=√(a−b)2
⇒ d=a−b
Length of perpendicular from C on the y− axis.
p=a−bn−m
∴ Area of the triangle fromed by the lines L1,L2 and y−axis
A=∣∣∣12dp∣∣∣
⇒ A=12×(a−b)×a−bn−m
⇒ A=(a−b)22(m−n)