mohon bantuannya, persamaan garis

tegak lurus garis 2x-y+3=0
m=-1/2

pusat=(-2/-2,-4/-2)
pusat = (1,2)

y-y1=m(x-x1)
y-2 =-1/2(x-1)
y-2=-1/2x+1/2
2y-4= -x+1 (C)
2y+x-5=0