x

y

4. 對稱

(2,3)

(2,-3)

(-2,3)

A

B

設點A(3,-1)，\;試求點A關於直線
\L:2x-3y+4=0的對稱點坐標。

(2,0)

(0,3)

A(3,-1)

A'(h,k)

M(\frac{h+3}{2},\frac{k-1}{2})

L

A'(-1,5)

H

H(1,2)

坐標平面上，\;假設一點光源位於點
\A(-1,2)，\; 朝向x軸射出，\;已知經x軸
\反射後，\;反射光通過B(5,3)，\;求x軸
\上的入射點位置。

A(-1,2)

B(5,3)

A'(-1,-2)

(\frac{7}{5},0)

坐標平面上，\;假設一點光源位於點A(2,-5)
\朝向直線L:y=x射出，\;已知經L反射後，\;
\反射光線通過點B(3,-2)，\;求直線L上的入
\射點坐標。

A(2,-5)

B(3,-2)

A'(-5,2)

A'(h,k)

M

Forward

M

s = 0 (slider from 0 to 57, step 1)
A = Corner(4)
B = (14.86, -6.98)
poly_0, b, a, c, d = Polygon(B, B + (18, 0), B + (18, 12), B + (0, 12))
C = PointIn(poly_0)
O = DynamicCoordinates(C, floor(x(C)), floor(y(C)))
u = Vector((x(B), y(O)), (x(B) + 18, y(O)))
v = Vector((x(O), y(B)), (x(O), y(B) + 12))
A_1: O + (2, 3)
t2 = 0.06 (slider from 0 to 1, step 0.01)
t3 = 0.05 (slider from 0 to 1, step 0.01)
A_2 = Translate(A_1, Vector[(t2 * Vector[A_1, A_1 - (0, 6)])])
w_1 = Vector(A_1, A_2)
A_3 = Translate(A_1, Vector[(t3 * Vector[A_1, A_1 - (4, 0)])])
w_2 = Vector(A_1, A_3)
P_1 = PointIn(poly_0)
P_2 = PointIn(poly_0)
Q_1 = DynamicCoordinates(P_1, floor(x(P_1)), floor(y(P_1)))
Q_2 = DynamicCoordinates(P_2, floor(x(P_2)), floor(y(P_2)))
d_0 = Segment(P_1, P_2)
e_0 = LineBisector(d_0)
α = Angle(e_0, d_0)
M_1: O + (2, 0)
M_2: O + (0, 3)
text1: "若直線L為線段\overline{AB}的中垂線，
\\點\;M\;為線段\overline{AB}的中點，\;則點B
\\稱為點A關於直線L的對稱點。\;
\\也稱A與B互為關於L的對稱點。
\\點\;M\;稱為點A關於直線L的投影
\\點或垂足。\;" + (LaTeX[If[s ≟ 2, "例如：\;已知點P(2,3) \\點\;P\;關於x軸的投影點為(2,0)，\\對稱點為(2,-3)； "]]) + "" + (LaTeX[If[s ≟ 3, "\\點\;P\;關於y軸的投影點為(0,3)，\\對稱點為(-2,3)"]]) + ""
B_1: O + (3, -1)
L_4: (2 * (x - x(O))) - (3 * (y - y(O))) = (-4)
text2: "" + (LaTeX[If[(s ≥ 5) ∧ (s ≤ 9), "法一 \\設A關於直線L的對稱點為A'(h,k)"]]) + "
\\" + (LaTeX[If[(s ≥ 6) ∧ (s ≤ 9), "則由\overline{AA'}⊥L知，\;m_L×m_{\overline{AA'}}=-1"]]) + "
\\" + (LaTeX[If[(s ≥ 7) ∧ (s ≤ 9), "又L:y=\frac{2}{3}x+\frac{4}{3}⇒m_L=\frac{2}{3}"]]) + "
\\" + (LaTeX[If[s ≟ 8 ∨ s ≟ 9, "m_{\overline{AA'}}=\frac{k+1}{h-3}=-\frac{3}{2}"]]) + "
\\" + (LaTeX[If[(s ≥ 9) ∧ (s ≤ 15), "⇒3h+2k=7"]]) + "
\\" + (LaTeX[If[(s ≥ 11) ∧ (s ≤ 15), "取\overline{AA'}的中點M(\frac{h+3}{2},\frac{k-1}{2})"]]) + "
\\" + (LaTeX[If[(s ≥ 12) ∧ (s ≤ 15), "又M在L上"]]) + "
\\" + (LaTeX[If[(s ≥ 13) ∧ (s ≤ 15), "\\⇒2\frac{h+3}{2}-3\frac{k-1}{2}+4=0"]]) + "
\\" + (LaTeX[If[(s ≥ 14) ∧ (s ≤ 15), "⇒2h-3k=-17"]]) + "
\\" + (LaTeX[If[s ≥ 15, "\left\{ 3h+2k=7\\ 2h-3k=-17\right.\ "]]) + "
\\" + (LaTeX[If[s ≥ 17, "解得h=-1,k=5"]]) + "" + (LaTeX[If[s ≥ 18, "\\即A'(-1,5)"]]) + ""
V_1 = (6.98, 1.64)
V_2: V_1 + (0.8, 0)
v_3 = Segment(V_1, V_2)
V_3: V_1 + (0.6, 0.2)
V_4: V_1 + (0.6, 0)
Poly_V, v_4, v_2, v_1 = Polygon(V_2, V_3, V_4)
B_2 = Mirror(B_1, L_4)
d_1 = Segment(B_1, B_2)
β = Angle(d_1, L_4)
M_B = Midpoint(B_1, B_2)
D_{20} = Point(L_4)
text3: "法二：
\\" + (LaTeX[If[s ≥ 20, "因為直線L的法向量為\fgcolor{yellow}{\;n\;=(2,-3)}"]]) + "
\\" + (LaTeX[If[s ≥ 21, "故設A關於L的投影點坐標為"]]) + "
\\" + (LaTeX[If[s ≥ 22, "H(3+2t,-1-3t),t∈ℝ"]]) + "
\\" + (LaTeX[If[s ≥ 23, "又H在直線L上"]]) + "
\\" + (LaTeX[If[s ≥ 24, "故2(3+2t)-3(-1-3t)+4=0"]]) + "
\\" + (LaTeX[If[s ≥ 25, "⇒13t+13=0⇒t=-1"]]) + "
\\" + (LaTeX[If[s ≥ 26, "⇒H(1,2)"]]) + "
\\" + (LaTeX[If[s ≥ 27, "故可得對稱點坐標為A'(-1,5)"]]) + ""
H: O + (1, 2)
e_1 = Segment(B_1, H)
γ = Angle(D_{20}, H, B_1)
δ = Angle(D_{20} + (2, (-3)), D_{20}, D_{20} + (3, 2))
C_A: O + (-1, 2)
C_B: O + (5, 3)
D = Point(u)
d_2: x = x(D)
E = Mirror(C_A, d_2)
D' = Dilate(D, -2, C_A)
D'' = Mirror(D', d_2)
w_{31} = Vector(C_A, D)
w_{32} = Vector(D, D'')
ε = Angle(C_A, D, D + (0, 1))
ζ = Angle(E, D, D + (0, 1))
C_C: O + (-1, (-2))
e = Ray(C_C, C_B)
E_A: O + (2, (-5))
E_B: O + (3, (-2))
L_{40}: y - y(O) = x - x(O)
E_C: O + ((-5), 2)
text41: "" + (LaTeX[If[s ≥ 41, "設點A(2,-5)關於L的對稱點為A'(h,k)"]]) + "
\\" + (LaTeX[If[s ≥ 42, "因為\overleftrightarrow{AA'}⊥L，\;故"]]) + "" + (LaTeX[If[s ≥ 43, "\fgcolor{yellow}{m_{\overleftrightarrow{AA'}}m_L=-1}"]]) + "" + (LaTeX[If[s ≥ 44, "又m_L=1"]]) + "
\\" + (LaTeX[If[s ≥ 45, "故m_{\overleftrightarrow{AA'}}=\frac{k+5}{h-2}=-1"]]) + "" + (LaTeX[If[s ≥ 46, "⇒h+k=-3"]]) + "
\\" + (LaTeX[If[s ≥ 47, "取\overline{AA'}之中點M(\frac{h+2}{2},\frac{k-5}{2})"]]) + "
\\" + (LaTeX[If[s ≥ 48, "因為M在L上，\;故\frac{h+2}{2}=\frac{k-5}{2}"]]) + "
\\" + (LaTeX[If[s ≥ 49, "⇒h-k=-7"]]) + "" + (LaTeX[If[s ≥ 50, "⇒h=-5,k=2"]]) + "" + (LaTeX[If[s ≥ 51, "即A'(-5,2)"]]) + ""
text52: "找\overleftrightarrow{A'B}與直線L的交點即為所求
\\" + (LaTeX[If[s ≥ 53, "\overleftrightarrow{A'B}:\frac{y+2}{x-3}=\frac{2+2}{-5-3}"]]) + "
\\" + (LaTeX[If[s ≥ 54, "⇒x+2y=-1"]]) + "
\\" + (LaTeX[If[s ≥ 55, " \left\{ x-y=0\\x+2y=-1 \right.\ "]]) + "" + (LaTeX[If[s ≥ 56, "⇒x=y=-\frac{1}{3}"]]) + "
\\" + (LaTeX[If[s ≥ 57, "即入射點坐標為(-\frac{1}{3},-\frac{1}{3})"]]) + ""
f_1 = Segment(E_A, E_C)
M_E = Midpoint(E_A, E_C)
η = Angle(O, M_E, E_C)
L_{53} = Line(E_B, E_C)
E_D: O - (1 / 3, 1 / 3)
P_M = Midpoint(P_1, P_2)
f = OrthogonalLine(D_{20}, L_4)
F = Point(f)
text35: "由平面鏡成像原理知，\;將點光源A(-1,2)
\\視為對平面鏡(x軸)的對稱點A'(-1,-2)
\\射出，\;穿越平面鏡直射到點B(5,3)
\\" + (LaTeX[If[s ≥ 36, "\overleftrightarrow{A'B}:\frac{y-3}{x-5}=\frac{-2-3}{-1-5}"]]) + "" + (LaTeX[If[s ≥ 37, "⇒5x-6y=7"]]) + "
\\" + (LaTeX[If[s ≥ 38, "y=0代入得x=\frac{7}{5}，\;"]]) + "" + (LaTeX[If[s ≥ 39, "\\故入射點坐標為(\frac{7}{5},0)"]]) + ""
g = OrthogonalLine(D_{20}, L_4)
G = Point(g)
w_3 = Vector(D_{20}, G)