Find the equation of the straight line with positive gradient and inclined at an angle of 45° to the line 3y-x+1=0 and passing through the point (2,0)?

##y=2x-4##
The given line is ##3y-x+1=0## ….(i)
Now, the general equation of a straight line is: ##y=mx+c##, where ‘m’ is the slope of the line with respect to the +ve direction of the x-axis.
Now,
The slope of the line (i) is ##1/3##
Let the required line be ##y=mx+c## ….(ii)
The difference in the slopes of lines (i) and (ii) is 45°
Hence,
##(m-1/3)/(1+m.(1/3)) = tan(45)##
##=> m-1/3 = 1+m/3##
##=> 3m – 1 = 3+m##
##=>2m = 4##
##=>m = 2##
Also, the line (ii) passes through the point (2,0).
So,
##y=mx+c##
## => 0 = 2m+c##
## =>c=-4##
Hence, the required line is:
##y=2x-4##