標題:

中四附加數(直線方程)一問?急急，就測驗

發問:

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。

最佳解答:

As follow AS~~~ 圖片參考：http://www.photo-host.org/img/450031screenhunter_03_oct._16_18.52.gif 圖片參考：http://www.photo-host.org/img/893274screenhunter_04_oct._16_18.52.gif

其他解答:

(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.

aa.jpg

 

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