persamaan garis yg melalui titik (5,-6) dan tegak lurus dengan garis 3y-5x+12=0

3y+5x+12 = 0
3y = -5x-12+0
3y = -5x+12
y = -5x+12/3
y = -5x+4
m1 = -5
m1×m2 = -1
-5×m2 = -1
m2 = -1/-5
m2 = 1/5
y-y1 = m2 (x-x1)
y+6 = 1/5 (x-5)
5(y+6) = 5[1/5(x-5)]
5y+30 = x-5
5y = x-5-30
5y = x-25
y = x-25/5
y = x-5
atau
5y+x = -25
atau
5y+x+25 = 0
#semoga bermanfaat