Equation of one of the line through the point 1,5 and making an angle of 45°  with the linex−3y+5=0 is ax+by+c=0 . If  a>0,b>0 then the value of a+b+c is

The equation of any line which makes an angle  α with the line ax+by+c=0  and passing through the point x1,y1 is y−y1=tanθ±αx−x1 here   Here  tanθ=−ab and   α=45°Equations of the required lines are  y−5=tanθ±45°x−1It implies   y−5=tanθ+45°x−1=1+tanθ1−tanθx−1 Substitute tanθ=13y−5=1+131−13x−1=2x−1y−5=2x−22x−y+3=0For the other line equationy−5=tanθ−45°x−1=tanθ−11+tanθx−1Substitute tanθ=13y−5=13−11+13x−1x+2y−11=0Since a>0,b>0, the sum of the values of a,b,c  is  1+2−11=−8