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##