let the slope of the required line is m
the slope of 5x-y+4=0 is 5
so
tan45=|(m-5)/(1+5m)|
(m-5)/(1+5m)=1 or (m-5)/(1+5m)=-1
m-5=1+5m or m-5=-1-5m
m=-3/2 or m=2/3
when m=-3/2
the line is
y-6=(-3/2)(x-2)
2y-12=-3x+6
3x+2y-18=0
when m=2/3
y-6=(2/3)(x-2)
3y-18=2x-4
2x-3y+14=0