AMATHS; equations of striaght lines

Let A(x1,y1) and C(x2,y2)
Length of OA = √(x1² + y1²) = 3
y1 = √(9 – x1²)
Sub into equation of OA,
x1 - √3√(9 – x1²) = 0
x1² = 3(9 – x1²)
4x1² = 27
x1 = 3√3 / 2 (-3√3 / 2 rejected as A is in the 1st quadrant)
y1 = √(9 – 27/4) = 3 / 2

A(3√3 / 2, 3/2)

Slope of OA = tan 30º = 1/√3
Slope of OA × Slope of AB = -1 (perpendicular lines)
Slope of AB = -1 / slope of OA = -1/(1/√3) = -√3

By point-slope form, equation of AB:
y – 3/2 = -√3 (x - 3√3/2)
2y – 3 = -2√3 x + 9
2√3 x – 2y + 12 = 0
√3 x – y + 6 = 0

Similarly,

Length of OC = √(x2² + y2²) = 3, y2 = √(9 – x2²)
y2 = √(9 – x2²)
Sub into equation of OC,
√3x2 + √(9 – x2²) = 0
3x2² = (9 – x2²)
4x2² = 9
x2 = -3/2 (3 / 2 rejected as C is in the 2st quadrant)
y2 = √(9 – 9/4) = 3√3 / 2

A(-3/2, 3√3 / 2)

Slope of BC = Slope of OA = 1/√3 (parallel lines)

By point-slope form, equation of BC:
y – 3√3/2 = (1/√3) (x + 3/2)
2√3y – 9 = 2 x + 3
2 x - 2√3y + 12 = 0
x - √3y + 6 = 0

2007-04-18 01:56:30 補充：
Sorry, 7th line from the bottom should be C(-3/2, 3√3 / 2) not A

Let (a,b) be the coordinate of A
a/3= cos 30
a= 3cos 30 =3 x sqrt3/ 2
b/3= sin 30
b= 3/2
Let (c,d) be the coordinate of C
c/3 = cos 120
c= -3/2
d/3= sin 120
d= 3 x sqrt3/ 2
Let (e,f) be the coordinate of B
Let a point X on the y-axis and AX is parallel to x-axis
angle XAO= 30 (alt angles, XA//x-axis)
angle BAX = 90-30=60
the horizontal distance from B to A is 3cos 60 = 3/2
e= 3 x sqrt3/ 2 - 3/2
= 3(sqrt3 -1)/2
the vertical distance from B to A is 3sin 60 = 3 x sqrt3/ 2
f= 3/2 - 3 x sqrt3/ 2
= 3(1-sqrt3) /2

slope of BC = (3 x sqrt3/ 2 - 3(1-sqrt3) /2)/(-3/2 - 3(sqrt3 -1)/2)
= (3/2)/ 3sqrt3/ 2
= 1/ sqrt 3

so equation of BC is (y- (3 x sqrt3/ 2))/ (x - (-3/2)) = 1/sqrt3
sqrt3 x y - 9 /2 = x +3/2
x - (root3) (y) +6 =0

because BC is perpendicular to BA, so slope of AB = -1/ (1/ sqrt 3)
= -sqrt 3
the equation of AB is (y- 3/2)/(x- (3 x sqrt3/ 2)) = -sqrt 3
y-3/2 = -sqrt3(x) +9/2
(root3)(x) +y -6 =0