math help!!!!!!!!

L1：4x-ky+3=0
ky = 4x+3
y = 4x/k +3/k
slope of L1 = 4/k

L2：(3x/k)+(y/3)=1
y/3 = -3x/k + 1
y = -9x/k + 3
slope of L2 = -9/k

As L1 and L2 is mutually perpendicular, the product of the slopes equals -1, i.e. -1 = (4/k)(-9/k)
1 = 36 / k²
k² = 36
k = 6 or -6

As slope of L1 is positive, k=6

∵直線L₁:4x-ky+3=0與 L₂:(3x/k)+(y/3)=1互相垂直
∴（L₁斜率）（L₂斜率）=-1
即(4/k)[-(3/k)/(1/3)]=-1
(4/k)(-9/k)=-1
-36/k²=-1
k²=36
k=4或k=-4(捨去)

2008-02-17 13:12:29 補充：
k=6或k=-6(捨去)

y= mx + c, m 是斜率

4x – ky + 3 = 0
變為
4x + 3 = ky
ky = 4x + 3
y = 4x/k + 3/k

L1的斜率是 4/k

3x/k + y/3 = 1
通分母後變為
9x + ky = 3k
ky = 3k – 9x
y = -9x/k + 3

L2的斜率是-9/k

由於L1與L2互相垂直，所以
(4/k)(-9/k) = -1
-36/k^2 = -1
k^2 = 36
k = +6 或 k = -6 (reject)