x

y

已知坐標平面上兩點A(6,1),B(2,0)，
\點P在L:x-2y+1=0上，\;若\overline{PA}+\overline{PB}
\有最小值m，\;試求\;P\;點坐標與m值。

B(2,0)

A(6,1)

已知坐標平面上兩點A(6,1),B(2,0)，\;點
\P\;在L:x-2y+1=0上，\;若|\overline{PA}-\overline{PB}|
\有最大值M，\;試求\;P\;點坐標與M值。

P

A'(h,k)

A'(4,5)

P

P(\frac{11}{4},\frac{15}{8})

M

根據平面鏡成像原理
\以及入射角等於反射角

有一道光線從第一象限沿著直線
\L:x-2y+3=0射向x軸上的\;P\;點，\;經
\x軸反射後，\;光線沿著另一直線L'離去，\;
\試求\;P\;點坐標與L'的方程式。

$θ$

$180°-θ$

L

L'

Forward

s = 0 (slider from 0 to 48, step 1)
A = Corner(4)
B = (14.8, -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))
L_1: x - x(O) - (2 * (y - y(O))) + 1 = 0
P_A: O + (6, 1)
P_B: O + (2, 0)
D = Point(L_1)
e_1 = Segment(D, P_A)
e_2 = Segment(D, P_B)
F = Mirror(P_A, L_1)
text4: "" + (LaTeX[If[(s ≥ 4) ∧ (s ≤ 13), "設A關於L的對稱點A'(h,k)"]]) + "
\\" + (LaTeX[If[(s ≥ 5) ∧ (s ≤ 13), "L:y=\frac{1}{2}x+\frac{1}{2}"]]) + "" + (LaTeX[If[(s ≥ 6) ∧ (s ≤ 13), "⇒m_L=\frac{1}{2}"]]) + "
\\" + (LaTeX[If[(s ≥ 7) ∧ (s ≤ 13), "因為L⊥\overline{AA'}"]]) + "" + (LaTeX[If[(s ≥ 8) ∧ (s ≤ 13), "⇒m_L×m_{\overline{AA'}}=-1"]]) + "
\\" + (LaTeX[If[(s ≥ 9) ∧ (s ≤ 13), "m_{\overline{AA'}}=\frac{k-1}{h-6}=-2"]]) + "" + (LaTeX[If[(s ≥ 10) ∧ (s ≤ 13), "⇒2h+k=-13"]]) + "
\\" + (LaTeX[If[(s ≥ 11) ∧ (s ≤ 13), "取\overline{AA'}的中點M(\frac{h+6}{2},\frac{k+1}{2})且M在L上"]]) + "
\\" + (LaTeX[If[(s ≥ 12) ∧ (s ≤ 13), "\frac{h+6}{2}-2\frac{k+1}{2}+1=0"]]) + "" + (LaTeX[If[s ≟ 13, "⇒h-2k=-6"]]) + "
\\" + (LaTeX[If[s ≥ 14, "\left\{ 2h+k=13\\ h-2k=-6\right."]]) + "" + (LaTeX[If[s ≥ 15, "⇒h=4,k=5"]]) + "" + (LaTeX[If[s ≥ 16, "即A'(4,5)"]]) + "
\\" + (LaTeX[If[s ≥ 17, "\overleftrightarrow{A'B}:\frac{y-0}{x-2}=\frac{5-0}{4-2}"]]) + "" + (LaTeX[If[s ≥ 18, "⇒5x-2y=10"]]) + "
\\" + (LaTeX[If[s ≥ 19, "\left\{ x-2y=-1\\5x-2y=10 \right. "]]) + "" + (LaTeX[If[s ≥ 20, "⇒x=\frac{11}{4},y=\frac{15}{8}"]]) + "
\\" + (LaTeX[If[s ≥ 21, "即P(\frac{11}{4},\frac{15}{8})"]]) + ""
L_{17} = Line(F, P_B)
P_P = Intersect(L_1, L_{17})
M_{11} = Midpoint(P_A, F)
e_8 = Segment(P_A, F)
e_{22} = Segment(P_P, P_A)
text22: "∵\Delta AMP≅\Delta A'MP\;(\;RHS\;)
\\" + (LaTeX[If[s ≥ 23, "∴\overline{PA}+\overline{PB}=\overline{PA'}+\overline{PB}"]]) + "
\\" + (LaTeX[If[s ≥ 24, "當A',P,B共線時， \\\overline{PA}+\overline{PB}有最小值\overline{A'B}"]]) + "
\\" + (LaTeX[If[s ≟ 25, "⇒m=\sqrt{2^2+5^2}=\sqrt{29}"]]) + ""
t_{try}: "" + (LaTeX[If[s ≟ 3, "\overline{PA}+\overline{PB}=" + (e_1 + e_2)]]) + "" + (LaTeX[If[s ≟ 29, "|\overline{PA}-\overline{PB}|=" + (abs(e_1 - e_2))]]) + ""
e_{30} = Line(P_A, P_B)
P_{32} = Intersect(L_1, e_{30})
t_{P32}: "P" + (LaTeX[If[s ≥ 34, "(-4,-\frac{3}{2})"]]) + ""
text30: "" + (LaTeX[If[(s ≥ 30) ∧ (s ≤ 34), "考慮\overleftrightarrow{AB}:\frac{y-0}{x-2}=\frac{1-0}{6-2}"]]) + "
\\" + (LaTeX[If[(s ≥ 31) ∧ (s ≤ 34), "⇒x-4y=2"]]) + "
\\" + (LaTeX[If[(s ≥ 32) ∧ (s ≤ 34), "⇒\left\{ x-2y=-1\\x-4y=2 \right."]]) + "
\\" + (LaTeX[If[(s ≥ 33) ∧ (s ≤ 34), "⇒兩式相減得\;y=-\frac{3}{2},x=-4"]]) + "
\\" + (LaTeX[If[s ≥ 34, "即P(-4,-\frac{3}{2})"]]) + "
\\" + (LaTeX[If[s ≥ 36, "此時|\overline{PA}-\overline{PB}|有最大值\overline{AB}"]]) + "
\\" + (LaTeX[If[s ≥ 37, "M=\sqrt{4^2+1^2}=\sqrt{17}"]]) + ""
L_{38}: x - x(O) - (2 * (y - y(O))) = (-3)
P_{39}: O + ((-3), 0)
text1: "P" + (LaTeX[If[s ≥ 40, "(-3,0)"]]) + ""
α = Angle(O, P_{39}, O + (-1, 1))
L_{44}: x = x(P_{39})
text39: "" + (LaTeX[If[s ≟ 40 ∨ s ≟ 39, "P\;點即L與x軸的交點"]]) + "
\\" + (LaTeX[If[s ≟ 40, "y=0代入L，\;得x=-3，\;即P(-3,0)"]]) + "
\\" + (LaTeX[If[s ≥ 41, "又L:y=\frac{1}{2}x+\frac{3}{2}，\;"]]) + "" + (LaTeX[If[s ≥ 42, "設L的斜角為θ，\;"]]) + "
\\" + (LaTeX[If[s ≥ 43, "故L之斜率為tanθ=\frac{1}{2}"]]) + "
\\" + (LaTeX[If[s ≟ 44 ∨ s ≟ 45, "因為\fgcolor{yellow}{入射角等於反射角}"]]) + "
\\" + (LaTeX[If[s ≥ 45, "所以L'的斜角為180°-θ"]]) + "
\\" + (LaTeX[If[s ≥ 46, "m_{L'}=tan(180°-θ)=-tanθ=-\frac{1}{2}"]]) + "
\\" + (LaTeX[If[s ≥ 47, "故L'的方程式為y=-\frac{1}{2}(x+3)"]]) + "
\\" + (LaTeX[If[s ≥ 48, "⇒L':x+2y+3=0"]]) + ""
L_{45}: x - x(O) + (2 * (y - y(O))) + 3 = 0
β = Angle(O, P_{39}, O + ((-5), 1))
t = 2 (slider from 0 to 2, step 0.01)
Q_1: P_{39} + (10, 5)
Q_2: P_{39} + ((-10), 5)
Q_3 = Translate(Q_1, Vector[(t * Vector[Q_1, P_{39}])])
Q_4 = Translate(P_{39}, Vector[((t - 1) * Vector[P_{39}, Q_2])])
q_1 = Vector(Q_1, Q_3)
q_2 = Segment(P_{39}, Q_1)
q_3 = Vector(P_{39}, Q_4)
γ = Angle(P_{39} + (0, 1), P_{39}, P_{39} + (2, 1))
δ = Angle(P_{39} + (0, 1), P_{39}, P_{39} + ((-2), 1))