Question

Jika garis ax + 2y - 9 = 0 tegak lurus dengan garis A ( x + y ) - 6y + 4 = 0. Tentukanlah nilai a !

Answer 1

Ax + 2y - 9 = 0
2y = -ax + 9
y = -a/2 x + 9/2      (gradien koefisien x)
m1 = - a/2
M2 = -1/m1 = -1 / -a/2 = 2/a

a(x+y) - 6y + 4 = 0
ax+ ay - 6y + 4 = 0
ax + 4 = 6y -ay
ax + 4 = y (6 - a)
(ax + 4) / (6 - a) = y
(a / (6-a) ) x + 4 / (6 - a) = y
m2 = a/(6 - a)

2/a = a/ (6 -a)
12 - 2a = a²
0 = a² + 2a -12
a = (-2 + √(4 -4(1)(-12)) / 2
a = (-2 + √4 +48) / 2
a = (-2  + √ 52) / 2
a = (-2 + 2√13) / 2
a = -1 + √13    atau  a = -1 - √13