1.persamaan garis lurus yang melalui titik (2,-5) dan tegak lurus terhadap garis y-3x-1=0.....

gradien (m) = [tex] \frac{3}{1} [/tex] = 3
Karena Tegak lurus berarti gradiennya [tex]-\frac{1}{3} [/tex]

(2,-5)
x1 = 2
y1 = -5

y-y1 = m (x-x1)
y+5 = [tex]-\frac{1}{3} [/tex](x-2)
y = [tex] -\frac{1}{3}x[/tex] + [tex] \frac{2}{3} [/tex] - 5
-----------------------------------------------------------------------------x3
3y = -x + 2 - 15
3y = -x - 13

Bisa kayak gini :
1. 3y = -x -13
2. 3y + x + 13 = 0
3. 3y + x = -13
Tergantung options

bx - ay = bx1 - ay1 
1x - 3y = 1 (2) - (-3) (-5)
1x - 3y = 2- (-15)
1x - 3y = 2 + 15
1x - 3y = 17 
 x - 3y = 17
itu jawabannya...ada pilihannya ga?