2 F4 Amaths questions

1. Since the straight line L passes through the point of intersection of the lines L1: 3x - 4y + 7 = 0 and L2: x - 3y + 2 = 0. So, let

the equation of line L:

3x - 4y + 7 + k(x - 3y + 2) = 0

(3 + k)x - (4 + 3k)y + (7 + 2k) = 0 ───── (*), where k is a constant.


Let m be the slope of line L. M be the slope of L3: y = 2x + 1

m = -[-(3 + k)/(4 + 3k)] = (3 + k)/(4 + 3k)

M = 2

Since the line L makes an acute angle of 45° with L3.

So, tan45° = │(M - m)/(1 + Mm)│

1 = │[2 - (3 + k)/(4 + 3k)]/{1 + 2[(3 + k)/(4 + 3k)]}│

1 = │[(5 - 5k)/(4 + 3k)]/[(10 + 5k)/(3 + k)]│

1 = │5(1 - k)/5(2 + k)│

2 + k = 1 - k or 2 + k = -1 + k

k = -1/2 or 3 = 0 (no solution)

Therefore, the equation of L:

(3 - 1/2)x - [4 + 3(-1/2)]y + [7 + 2(-1/2)] = 0

5x/2 -5y/2 + 5 = 0

x - y + 2 = 0



2. Mid point of AC = [(m + 1) + 1]/2 , [2 + (-1)]/2 = (m + 2)/2 , 1/2

Mid point of BD = (-2 + 3)/2 , [(-2) + (n + 3)]/2 = 1/2 , (n + 1)/2


If ABCD is a parallelogram, then the mid-point AC and that of BD should be the same point.

i.e. (m + 2)/2 = 1/2

m = -1

1/2 = (n + 1)/2

n = 0


2007-06-18 22:51:39 補充：
Amendment:So, tan45° = │(M - m)/(1 + Mm)│1 = │[2 - (3 + k)/(4 + 3k)]/{1 + 2[(3 + k)/(4 + 3k)]}│1 = │[(5 + 5k)/(4 + 3k)]/[(10 + 5k)/(3 + k)]│1 = │5(1 + k)/5(2 + k)│2 + k = 1 + k or 2 + k = -1 - kk = -3/2 or 1 = 0 (no solution)

2007-06-18 22:52:52 補充：
Therefore, the equation of L:(3 - 3/2)x - [4 + 3(-3/2)]y + [7 + 2(-3/2)] = 03x/2 + y/2 + 4 = 03x + y + 8 = 0

2007-06-19 18:22:43 補充：
任何經過L1和L2相交的那一點的直線都可設七的方程為3x - 4y + 7 + k(x - 3y + 2) = 0。因為L satisfy呢個條件，所以佢自然可以設為這方程。

3x - 4y + 7 + k(x - 3y + 2) = 0 ------- (1)
(3 + k)x - (4 + 3k)y + 7 + 2k = 0
slope = (3 + k) / (4 + 3k)

slope of L3 = 2

! (3 + k) / (4 + 3k) - 2 / 1 + 2(3 + k) / (4 + 3k) ! = tan45 = 1
! (k + 1) / (k + 2) ! = 1 ---------------- (2)
k = -3/2

sub it into (1), eqt. of L
3x + y + 8 = 0

呢題小心比佢暗到, L2 : x - 3y + 2 = 0 都同 L3 成 45 度,
因為通常都會搵到兩條線同 L3 做到 45 度, (2) 果度得一個 k 值,
你就要 check 下 L1 同 L2
不過 L2 算唔算係答案,就見仁見智

2)
mid-point of AC = ( (m+2)/2, 1/2 )
mid-point of BD = ( 1/2, (n+1)/2 )

ABCD is a parallelogram, so mid-point of AC = mid-point of BD
(m+2)/2 = 1/2
m = -1

1/2 = (n+1)/2
n = 0

2007-06-18 21:19:40 補充：
上面老兄好似做錯少少野