Answer:
4x-y+1=0
Step-by-step explanation:
here,given equation of a line id
4x-y-2=0.. eqn(i)
equation of any line parallel to line (i) is
4x-y+k=0...eqn(ii)
since, the line(ii) passes through (1,5)[replacing x=1 and y=5 in eqn(ii), we get]
4*1-5+k=0
or, 4-5+k=0
or,-1+k=0
•°•k=1
substituting the value of k=1 in eqn(ii),
4x-y+1=0 is the required equation of the line.
