Yes, we can construct an equilateral triangle ABC with the given conditions using rotations. Starting from any one of the three parallel lines and rotating the other two lines by 60 degrees each will yield the other two points of the triangle, allowing its construction.
6. 1. ROTATIONJ FORMULA OVER O(0,0) .
Let A(x,y) any point in plane V ang A’(x’,y’)is
image of A over R 0, , or A’ = R 0, (A).
Let m(XOA)= .
We have x =OA cos dan y = OA sin and
x’ = OA’ cos (+)
= OA (cos cos - sin sin)
= x cos - y sin
8. .
• y’ = OA’ sin (+)
• = OA(sin cos + cos sin )
• = x sin + y cos
• so
• x’ = xcos - y sin
• y’ = x sin + y cos
• or
x' cos sin x
y ' sin y
cos
10. 2. ROTATION OVER P(a,b)
• Let we have coordinate system with
centre P(a,b)and has two axis X*
and Y*, X//X* and Y//Y*.
• If C(x*,y*) and C’=RP,(C), then C’
(x*’,y*’) , we have a relation :
x*' cos sin x *
y*' sin
cos y *
11. In coordinate of X , Y axis , we have :
x'a cos sin x a
y 'b sin cos y b
x' cos sin x p
y ' sin cos y q
p a cos b sin a
q a sin b cos b
12. THEOREM
• Rotation RP, can represent in
composition of two lines
reflection over s and t with P is
(s,t) and m(<(s,t))=½ .
• Rotation is an isometry
• composition of two lines
reflection :
SAB ,if s//t
M t Ms
R P,θ , if t and s not paralel
23. • a. Fixed any point A on a.
• b. Rotated line c, with angle 60o over A, we got c’.
• c. Intersection of line c’ and line b, ( c’,b) is point
B.
• d. We can construct equilateral triangle ABC.
•
• We can also start with fixed point B on b or C on
c.
• Can do it ?
24. • a. Fixed any point B on b.
• b. Rotated line a, with angle 60o over B, we got a’.
• c. Intersect of line a’ and line c, ( a’,c) is point C .
• d. We can construct equilqteral triangle ABC.
•
• We can also start with fixed point A on a or C on
c.
• Can do it ?