diberikan garis l : (x-2y) + a(x+y) = 5 dan garis g : (5y-3x) -3a(x+y) = 12. tentukan nilai a agar : a. garis l sejajar g b. garis tegaklurus garis g

[tex]L: \\ (a+1)x+(a-2)y=5 \\ y= \frac{(a+1)}{(a-2)}x+ \frac{5}{(a-2)} \\ \\ G: \\ (-3-3a)x+(5-3a)y=12 \\ y= x+ \frac{12}{(5-3a)}[/tex]

a.
[tex]\frac{(a+1)}{(a-2)}=\frac{(3+3a)}{(5-3a)}\\ (a+1)(5-3a)=(3+3a)(a-2) \\ -3a^2+2a+5=3a^2+3a-12 \\ 6a^2+a-17=0[/tex]

b.
[tex](\frac{(a+1)}{(a-2)})(\frac{(3+3a)}{(5-3a)})=-1 \\ \frac{3a^2+6a+3}{-3a^2+11a-10} =-1 \\ 3a^2+6a+3=3a^2-11a+10 \\ 17a=7 \\ a= \frac{7}{17} [/tex]