Straight Line

1.
L1 : ax + by + c = 0 .. (Make a > 0)
slope of L1 = -a/b > 0 .. Hence, b < 0
x-intercept of L1 = -c/a > 0 .. Hence, c < 0
y-intercept of L1 = -c/b < 0 .. Hence, c < 0

L2 : px + qy + r = 0 .. (Make p > 0)
slope of L2 = -p/q < 0 .. Hence, q > 0
x-intercept of L2 = -r/p > 0 .. Hence, r < 0
y-intercept of L2 = -r/q > 0 .. Hence, r < 0

(slope of L1) = -a/b > 0 ..and..(slope of L1­) = -p/q < 0
Hence, -a/b > -p/q
a/b < p/q
b < 0 and q > 0, then bq < 0
Hence, (a/b)bq > (p/q)bq
aq > bp ...... (i) is false

x-intercept of L1 = x-intercept of L2
-c/a = -r/p
(-c/a)(-ap) = (-r/p)(-ap)
cp = ar ...... (ii) is false.

y-intercept of L1 = -c/b < 0 ..and..y-intercept of L2 = -r/q > 0
-c/b < -r/q
c/b > r/q
b < 0 and q > 0, then bq < 0
(c/b)(bq) < (r/q)(bp)
cq < br ...... (iii) is true.

Ans: Only (iii) is true.


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2.
Let (c, e) be the coordinates of C.
(Refer to the figure, c > 0, e > 0)

Length of side of the square = AD = √[(2 + 1)² + (-1- 3)²] = 5
Slope of AD = (-1 - 3) / (2 + 1) = -4/3

To find the coordinates of C(c, e) :
CD : √[(c + 1)² + (e - 3)²] =5 ...... [1]
Slope of CD : (e - 3) / (c + 1) = -1 / (-4/3) ...... [2]

From [2] :
(e - 3) / (c + 1) = 3/4
e - 3 = (3c + 3)/4
e = (3c + 15)/4 ...... [3]

Put [3] into [1] :
√{[c + 1]² + [(3c + 15)/4 - 3]²} =5
√{[c + 1]² + [(3c + 3)/4]²} =5
c² + 2c + 1 + [(9c² + 18c+ 9)/16] = 25
16c² + 32c + 16 + 9c² + 18c+ 9 = 400
25c² + 50c - 375 = 0
c² + 2c - 15 = 0
(c - 3)(c + 5) = 0
c = 3 or c = -5 (rejected, for c > 0)

Put c = 3 into [3] :
e = [3(3) + 15]/4
e = 6

Hence, coordinates of C = (3, 6)

Let (d, f) be the coordinates of D.

The diagonals of square bisect each other.

The coordinates of the mid-point of AC = The coordinates of the mid-point of BD
(2 + 3)/2 = (b - 1)/2 ..and.. (-1 + 6)/2 = (f + 3)/2
b - 1 = 5 ..and.. f + 3 = 5
b = 6 ..and.. f = 2

Hence, coordinates of B = (6, 2)

2015-03-14 01:28:42 補充：
1.
To make a > 0 and p > 0 (i.e. to make the coefficent of x > 0) :

For example, rewrite the equation -2x + 3y - 4 = 0 as 2x - 3y + 4 = 0

If L1 : -x + 2y + 3 = 0, and L2 : 2x + 3y - 6 = 0
then cq = 6 and br = -8, so (iii) is false.

(The question did not mention a and p must be positive)

第一題的 ii 是有點兒特別的。

你再看看圖像。

你留意到兩條線有相同的 x-intercept 嗎？

即是當 y = 0 的時候，兩線得出的 x 值相同。

因此，-c/a = -r/p

即 cp = ar

故 ii 不對。