考慮直線族L：(2x-y+8)+k(x-4y+11)=0
(a) 對於任何實數k，L都能通過一個固定點A。求A的坐標。
(b)試以k表示L的斜率。
(c)設L0為直線族中L的一條直線，其中k=0。求兩個於直線族L內的直線方程，其中該兩條直線與L0之間的夾角圴為45度。
中四附加數(直線方程)一問?急急，就測驗
2008-10-16 2:03 pm
回答 (2)
2008-10-17 2:57 am
2008-10-16 4:12 pm
(a) The fixed point A is the intersecting point of 2x-y + 8 = 0 and x-4y + 11 = 0.
Sub. y = 2x + 8 into the 2nd equation, we get x - 4(2x + 8 ) + 11 = 0
x - 8x - 32 + 11 = 0
7x = - 21
x = -3, y = 2(-3) + 8 = -6 + 8 = 2. So A is (-3, 2).
b) (2x - y + 8)+ k(x - 4y + 11) = 0
2x - y + 8 + kx - 4ky + 11k = 0
(2 +k)x - (1 + 4k)y + (11k + 8) = 0
Slope = (2+k)/(1+4k).
c)
L0 is 2x - y + 8 = 0, slope = 2.
So slope of L is
(2 + 1)/(1 - 2 x 1) = -3 = (2+k)/(1 +4k)
-3 - 12k = 2 +k
-5 = 13k
k = -5/13
So one of the line is
(2x - y +8) -5(x - 4y + 11)/13 = 0
26x - 13y + 104 - 5x + 20y - 55 = 0
21x + 7y + 49 = 0
3x + y + 7 = 0.
The other one is perpendicular to this one, so slope = 1/3 and passing through A.
So equation is
y - 2 = (x +3)/3
3y - 6 = x + 3
3y = x + 9.
Sub. y = 2x + 8 into the 2nd equation, we get x - 4(2x + 8 ) + 11 = 0
x - 8x - 32 + 11 = 0
7x = - 21
x = -3, y = 2(-3) + 8 = -6 + 8 = 2. So A is (-3, 2).
b) (2x - y + 8)+ k(x - 4y + 11) = 0
2x - y + 8 + kx - 4ky + 11k = 0
(2 +k)x - (1 + 4k)y + (11k + 8) = 0
Slope = (2+k)/(1+4k).
c)
L0 is 2x - y + 8 = 0, slope = 2.
So slope of L is
(2 + 1)/(1 - 2 x 1) = -3 = (2+k)/(1 +4k)
-3 - 12k = 2 +k
-5 = 13k
k = -5/13
So one of the line is
(2x - y +8) -5(x - 4y + 11)/13 = 0
26x - 13y + 104 - 5x + 20y - 55 = 0
21x + 7y + 49 = 0
3x + y + 7 = 0.
The other one is perpendicular to this one, so slope = 1/3 and passing through A.
So equation is
y - 2 = (x +3)/3
3y - 6 = x + 3
3y = x + 9.
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