國中模凝考 1題求解?

B點即為直線L的x截距:
令 y = 0
0 = kx + 9
x = - 9/k
故B點座標為 ( -9/k , 0 )

C點即為直線L的y截距:
令 x = 0
y = k*0 + 9 = 9
故C點座標為 ( 0 , 9 )

三角形BOC面積
= (1/2) * OC * OB
= (1/2) * 9 * ∣-9/k∣
= (81/2) * ∣1/k∣
= 81

∣1/k∣ = 2
k = ± 1/2
故滿足條件的直線L有兩種可能:
y = (1/2)x + 9 或 y = - (1/2)x + 9

當 k = 1/2
L : y = (1/2)x + 9
M : y = (1/2)x - 1
兩直線斜率相同,但y截距不同,
故兩直線相互平行,不可能有交點,與題意矛盾.

當 k = - 1/2
L : y = -(1/2)x + 9 ... (1)
M : y = (1/2)x - 1 ... (2)
(1) + (2) 得:
2y = 8
y = 4 , 代回(2)式得
4 = (1/2)x - 1
8 = x - 2
x = 10
即A點座標為 ( 10 , 4 )

Ans: ( 10 , 4 )