The equation of line ab is y = 5x + 1. write an equation of a line parallel to line ab in slope-intercept form that contains point (4, 5). y = 5x − 15 y = 5x + 15 y = x + y = x −

METHOD 1:
The equation of line parallel to ax+by=c is,
ax+by=k
Similarly,in this case equation of line parallel to the st line y=5x+1 (i.e, 5x-y=-1) is given by,
5x-y=k(i)
As equation (i) is the line containing (4,5),
k=5×4-5=15
Hence equation (i) becomes,
5x-y=15
y=5x-15➡required equation (ANS)

METHOD 2:

As slope of AB is 5, the slope of line parallel to AB should also be 5.
Thus, the equation of st line parallel to AB (with slope 5) & passing through (4,5) is:
m=(y-y1)/(x-x1)
5=(y-5)/(x-4)
5x-20=y-5
y=5x-15➡ required eqn

This does not make sense