diketahui persamaan garis g:6x+(5-2a)y-8=0 dan h:(a+1)x-2y+15=0. jika garis g tegak lurus garis h, nilai a=.....

Agar tegak lurus,
[tex]m_g\times m_h=-1 \\ -\frac{6}{5-2a}\times -\frac{a+1}{-2}=-1 \\ \frac{6}{5-2a}\times \frac{a+1}{2}=1 \\ \frac{6}{5-2a}=\frac{2}{a+1} \\ 6(a+1)=2(5-2a) \\ 6a+6=10-4a \\ 10a=4 \\ a=\frac{2}{5}[/tex]

Mg = - 6/ 5-2a
Mh = a+1 / 2

karena garis g dan h tegak lurus maka, Mg x Mh = -1
- 6 / 5-2a X a+1 /2 = -1
-6a -6 = -1
10 - 4a
-6a -6 = -10 + 4a
4 = 10a
a = 4/ 10 atau 2/5