✔ 最佳答案
Slope of line L1 = 2 = tan A
Slope of line L2 = 3 = tan B
A + m = B (where m stands for theta.)
tan m = tan (B -A) = (tan B - tan A)/(1 + tan A tan B) = (3 -2)/(1 + 6) = 1/7.
Let slope of the other line be tan C.
so C + m = A
tan C = tan (A - m) = (tan A - tan m)/(1 + tan A tan m) = (2 - 1/7)/(1 + 2/7)
= (13/7)/(9/7) = 13/9.
so the other line is y = 13x/9 or 9y = 13x.
Q2.
Let slope of BC be m,
so m[(-8 +3)/(3-2)] = -1
-5m = -1, m = 1/5.
so equation of BC is
y + 8 = (x -3)/5
5y + 40 = x - 3
5y = x - 43.
For point B, x - coordinate = -2 and it is on line BC,
so 5(-a) = -2 - 43 = -45
a = 9, so coordinates of B is (-2,-9).
Let coordinates of point C be (h,k) and it is on line BC,
so 5k = h - 43................(1)
Slope of AC x slope of AB = -1
[(k+3)/(h + 2)][(-9 + 3)/(-2 -2)] = -1
(k+3)/(h + 2) [-5/-4] = -1
5(k + 3) = -4(h +2)
5k + 15 = -4h - 8
5k + 4h = - 23.............(2)
Sub. (1) into (2) we get
5k + 4(5k + 43) = - 23
25k = -23 - 172 = -195
k = -39/5
h = 5k + 43 = -39 + 43 = 4
so C is (4, -39/5)
2009-04-17 15:57:06 補充：
Correction: slope of AC should be (k+3)/(h -2), not (k+3)/(h + 2). Slope of BC should be (-6/-4), not (-5/-4). So eqt (2) should be 3k + 2h = -5. Solving the 2 eqts, C should be (8,-7). Sorry for the mistake.
