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Verified answer
Jarak (0 , 0) ke y = -3x/a + 3 (y = mx + c)d = |y₁ - mx₁ - c| / √(1 + m²)
d = |0 - 0 - 3| / √(1 + [-3/a]²)
d = 3 / √(1 + 9/a²)
Jarak antara (a , 0) dan (0 , 3)
p = √[(0 - a)² + (3 - 0)²]
p = √(a² + 9)
Karena ½ p = d , maka
½ [√(a² + 9) ] = 3 / √(1 + 9/a²)
Kuadratkan Kedua Ruas...
¼ (a² + 9) = 9 / (1 + 9/a²)
¼ (a² + 9)(1 + 9/a²) = 9
a² + 9 + 9 + 81/a² = 9 • 4
a² + 81/a² + 18 = 36
a² + 81/a² - 18 = 0
(a - 9/a)² = 0
a = 9/a
a² = 9
a = ± 3
Jadi a = 3 atau a = -3