Proof of parallel lines

Assume both A and C are in the 3rd quadrant for ease of explanation.
Slope of AB = 2 = tan x ( x is the angle between line AB and the x - axis)
Slope of BC = 1/2 = tan y ( y is the angle between line BC and the x - axis).
So angle ABC = y - x.
Since triangle ABC is an isos. triangle, angle ACB = angle ABC = y - x.
Let angle between line AC and the x - axis be z.
so z = y + ( y - x) = 2y - x ( ext. angle of triangle ABC)
so tan z = tan ( 2y - x)
= (tan 2y - tan x)/(1 + tan 2y tan x)
= [2tan y/(1 - tan^2 y) - tan x]/[1 + 2 tan y tan x/(1 - tan^2 y)].
Sub. in the values of tan x = 2 and tan y = 1/2, we get tan z = - 2/11.
That means slope of AC = - 2/11, so AC is parallel to line 2x + 11y = 0.
(b)
Let the y - coordinates of point C be c, so x - coordinate = 2c - 1 since it is on line BC.
So slope of AC = (c - 2)/(2c - 1 - 3) = -2/11
11(c - 2) = -2(2c - 4)
11c - 22 = - 4c + 8
15c = 30
c = 2, so the coordinates of C is (3, 2)


2009-10-16 19:56:13 補充：
Correctio: Line 1, it should be 1st quadrant, not 3rd quadrant. Line 4 : angle ABC = x - y, not y - x = angle ACB. Line 7 : z = 180 - 2(x - y) + x = 180 - x + 2y = 180 + (2y - x), so tan z = tan [180 + (2y -x)] = tan (2y - x).

2009-10-16 22:25:28 補充：
Correction on part (b). Slope of AC = (c - 3)/(2c - 1- 2) = -2/11. 11(c - 3) = -2(2c - 3), 11c - 33 = -4c + 6, 15c = 39, so c = 13/5, coordinates of C should be (21/5, 13/5). Sorry for the mistake.