Author: 藍邦偉  老師

1.\; 斜率

直線方程式

A(x_1,y_1)

B(x_2,y_2)

A(x_1,y_1)

B(x_2,y_2)

y_2-y_1

x_2-x_1

x_2-x_1

y_1-y_2

$m$

1

m

1

m

A(x_1,y_1)

B(x_2,y_2)

水平，\;斜率m_{\overline{AB}}=0

直線斜率的正負決定傾斜的方向，\其斜率的絕對值決定傾斜的程度。

範例\
xy平面上，\;設m_1,m_2,m_3,m_4分別表L_1,L_2,L_3,L_4四直線之斜率，\;則下列何者正確？\
(A)m_1<m_2<m_3<m_4\;(B)m_4<m_3<m_2<m_1\;(C)m_3<m_4<m_1<m_2\;(D)m_4<m_3<m_1<m_2

L_1

L_2

L_3

L_4

範例\
xy平面上，\;設m_1,m_2,m_3,m_4分別表L_1,L_2,L_3,L_4四直線之斜率，\;則下列何者正確？\
(A)m_1＞m_2＞m_3＞m_4\;(B)m_4＞m_3＞m_2＞m_1\;(C)m_2>m_3>m_1>m_4\;(D)m_1>m_2>m_4>m_3\
(E)m_2>m_1>m_3>m_4

L_1

L_2

L_3

L_4

範例：\
試判別A(-13,13),B(2,7),C(26,-3)三點是否共線?

範例：\
一位海盜欲將三件珠寶埋藏在一個島上的三個地方，\;海盜就以島上的一棵大王椰子樹為\
中心，\;由大王椰子樹向東走12步埋他的第一件珠寶；\;由大王椰子樹向東走4步，\;再往北\
走a步埋他的第二件珠寶；最後\;由大王椰子樹向東走a步，\;再向南走8步埋他的第三件珠\
寶，\;事隔多年之後，\;海盜僅記得a>0及埋藏珠寶的三個地方在同一直線上，\;試求a值。

O

A(12,0)

B(4,a)

C(a,-8)

練習\
在坐標平面上任意畫一直線，\;並在直線上任意寫出三個點的\坐標。

A(1,1)

B(5,5)

C(3,3)

斜角與斜率

$α=%v$

Forward

x

y

x

y

x

y

x

y

x

y

B = (3.6, -6.08)
O_3 = DynamicCoordinates(B, floor(x(B)), floor(y(B)))
u_{x3} = Vector(O_3 - (2, 0), O_3 + (8, 0))
s = 0 (slider from 0 to 58, step 1)
A = Corner(4)
O_1 = (-0.98, -6.38)
O_2 = Point(y = y(O_1))
w_1 = 1 (slider from 0.7 to 2.5, step 0.1)
h_1 = 0.6 (slider from 0 to 1.5, step 0.1)
A_2: O_1 + (0, h_1)
A_3: O_1 + (w_1, h_1)
A_4: O_1 + (w_1, (2 * h_1))
A_5: A_4 + (w_1, 0)
A_6: A_5 + (0, h_1)
A_7: A_6 + (w_1, 0)
A_8: A_7 + (0, h_1)
A_9: A_8 + (w_1, 0)
A_1: (x(A_9), y(O_1))
poly_1, o_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, a_1, o_2 = Polygon(O_1, A_2, A_3, A_4, A_5, A_6, A_7, A_8, A_9, A_1, O_1)
w_2 = 1 (slider from 0.7 to 2.5, step 0.1)
h_2 = 0.6 (slider from 0 to 1.5, step 0.1)
B_2: O_2 + (0, (4 * h_2))
B_3: B_2 + (w_2, 0)
B_4: B_3 + (-(0, h_2))
B_5: B_4 + (w_2, 0)
B_6: B_5 + (-(0, h_2))
B_7: B_6 + (w_2, 0)
B_8: B_7 - (0, h_2)
B_9: B_8 + (w_2, 0)
B_1: O_2 + ((4 * w_2), 0)
poly_2, o_3, b_2, b_3, b_4, b_5, b_6, b_7, b_8, b_9, b_1, o_4 = Polygon(O_2, B_2, B_3, B_4, B_5, B_6, B_7, B_8, B_9, B_1, O_2)
text2: "" + (LaTeX[If[s ≟ 1, "問題：\;如圖所示，\;兩個階梯，\;如何判斷那一個坡度比較陡？"]]) + "\\" + (LaTeX[If[s ≟ 2, "如果將一個斜坡的程度量化，\;我們可以考慮以水平方向每\\前進一個單位時，\;鉛直方向上升或下降多少單位來表示。"]]) + ""
A_O: O_1 - (1, 1)
B_O: O_2 - (1, 1)
u_{y2} = Vector(B_O - (0, 1), B_O + (0, 8))
u_{y1} = Vector(A_O - (0, 1), A_O + (0, 8))
u_{x1} = Vector(A_O - (1, 0), A_O + (12, 0))
u_{x2} = Vector(B_O - (1, 0), B_O + (12, 0))
m_1 = Slope(Line[A_2, A_4])
m_2 = Slope(Line[B_2, B_4])
f_2(x) = If[x(B_O) - 1 ≤ x ≤ x(B_O) + 10, (m_2 * (x - x(B_3))) + y(B_3)]
f_1(x) = If[x(A_O) - 1 ≤ x ≤ x(A_O) + 10, (m_1 * (x - x(A_2))) + y(A_2)]
C_1 = Point(f_1)
C_2 = Point(f_1)
D_1 = Point(f_2)
D_2 = Point(f_2)
C_0: (x(C_2), y(C_1))
D_0: (x(D_2), y(D_1))
c_1 = Segment(C_1, C_0)
c_2 = Segment(C_2, C_0)
d_1 = Segment(D_1, D_0)
d_2 = Segment(D_0, D_2)
C_3: C_1 + (2, 0)
D_3: D_1 + (2, 0)
C_4: (x(C_3), f_1(x(C_3)))
D_4: (x(D_3), f_2(x(D_3)))
poly_{31}, d_6, d_8, d_7 = Polygon(C_1, C_4, C_3)
poly_{32}, d_3, d_4, d_5 = Polygon(D_1, D_4, D_3)
u_1 = Vector(O_3 - (2, 0), O_3 + (8, 0))
u_{y3} = Vector(O_3 - (0, 2), O_3 + (0, 6))
E_1 = (5.66, -5.56)
E_2 = (8.28, -3.32)
poly_0, a, b, c, d = Polygon(O_3 - (1.95, 2), O_3 + (8, (-2)), O_3 + (8, 5.95), O_3 + ((-1.95), 5.95))
f_3(x) = If[-10 ≤ x ≤ 20, (Slope[Line[E_1, E_2]] * (x - x(E_2))) + y(E_2)]
m_3 = Slope(Line[E_1, E_2])
text81: "" + (LaTeX[E_1 - O_3]) + ""
text82: "" + (LaTeX[E_2 - O_3]) + ""
text8: "練習：\\
(1)\;求過兩點(3,1)與(7,4)的直線斜率並畫出直線圖形。\\" + (LaTeX[If[s ≟ 10 ∨ s ≟ 11, "(2)\;求過兩點(-1,5)與(2,-1)的直線斜率並畫出直線圖形。"]]) + ""
text6: "m_{\overline{AB}}=" + (LaTeX[If[x(E_1) ≠ x(E_2), "" + (FractionText[m_3]) + "≈" + m_3, "沒有斜率"]]) + ""
text13: "" + (LaTeX[If[m_3 > 0, "m>0", If[m_3 ≟ 0, "m=0", If[m_3  <  0, "m<0", If[x(E_1) ≟ x(E_2), "沒有斜率"]]]]]) + ""
C = (1.26, -5.34)
O_4 = DynamicCoordinates(C, floor(x(C)), floor(y(C)))
poly_4, e, f, g, h = Polygon(O_4 - (3, 3), O_4 + (12, (-3)), O_4 + (12, 7.95), O_4 + ((-3), 7.95))
u_{y4} = Vector(O_4 - (0, 3), O_4 + (0, 8))
u_{x4} = Vector(O_4 - (3, 0), O_4 + (12, 0))
D = Point(u_{x4})
L_A: y - y(D) = (0.4 * (x - x(D)))
L_B: y - y(D) = (2.5 * (x - x(D)))
L_C: y - y(D) = ((-2) * (x - x(D)))
L_D: y - y(D) = ((-0.2) * (x - x(D)))
text24: "∵向右方上升的直線斜率為正\\" + (LaTeX[If[s ≥ 25, "\;\;\;向右方下降的直線斜率為負"]]) + "\\" + (LaTeX[If[s ≥ 26, "∴m_2>m_1>0>m_3>m_4"]]) + "\\" + (LaTeX[If[s ≥ 27, "故選(E)"]]) + ""
E = (3.6, -3.06)
M_A: y - y(E) = (0.5 * (x - x(E)))
M_C: y - y(E) = ((-0.4) * (x - x(E)))
F = (4.34, -5.24)
M_D: y - y(F) = (-(x - x(F)))
text29: "\overline{AB}的斜率為m_{\overline{AB}}=\frac{7-13}{2-(-13)}=-\frac{2}{5}\\" + (LaTeX[If[s ≥ 30, "\overline{BC}的斜率為m_{\overline{BC}}=\frac{-3-7}{26-2}=-\frac{5}{12}"]]) + "\\" + (LaTeX[If[s ≥ 31, "因為m_{\overline{AB}}≠m_{\overline{BC}}"]]) + "\\" + (LaTeX[If[s ≥ 32, "所以A,B,C不共線"]]) + ""
text34a: "設大王椰子樹所在位置為原點O(0,0),A(12,0),B(4,a),C(a,-8)\\" + (LaTeX[If[s ≥ 35, "∵A,B,C共線"]]) + "" + (LaTeX[If[s ≥ 36, "∴m_{\overline{AB}}=m_{\overline{AC}}"]]) + "\\" + (LaTeX[If[s ≥ 37, "⇒\frac{a-0}{4-12}=\frac{-8-0}{a-12}"]]) + "\\" + (LaTeX[If[s ≥ 38, "⇒a^2-12a=64"]]) + "\\" + (LaTeX[If[s ≥ 39, "⇒(a-16)(a+4)=0"]]) + "" + (LaTeX[If[s ≟ 40, "∴a=16"]]) + ""
O_5 = (-0.28, -5.88)
H = (-3.7, -8.3)
poly_5, h_3, i, j, k = Polygon(H, H + (12, 0), H + (12, 8), H + (0, 8))
u_{x5} = Vector(O_5 - (10, 0), (x(H) + 12, y(O_5)))
u_{y5} = Vector((x(O_5), y(H)), (x(O_5), y(H) + 8))
G_1: O_5 + (3, 0)
G_2: O_5 + (1, 4)
G_3: O_5 + (4, (-2))
a_G = Segment(G_2, G_3)
H_1: O_3 + (0, 1)
H_2: O_3 + (2, 3)
H_3: O_3 + (4, 5)
N_a = Line(H_1, H_2)
text41A: "探究活動\\
在坐標平面上，直線上每一個點P的坐標都是一組實數對(x,y)，\;其中x,y之間的代數關係性，\;便是平\\
面上直線的代數方程式。\\" + (LaTeX[If[s ≟ 42 ∨ s ≟ 43, "例如：\;A(0,1),B(2,3),C(4,5)三個點的y坐標都比x坐標大1，\;因此直線的代數方程式為y=x+1"]]) + "" + (LaTeX[If[s ≟ 44 ∨ s ≟ 45, "例如：\;D(2,1),E(4,2),F(6,3)三個點的y坐標都是x坐標的一半，\;因此直線的代數方程式為y=\frac{1}{2}x"]]) + ""
H_4: O_3 + (2, 1)
H_5: O_3 + (4, 2)
H_6: O_3 + (6, 3)
N_b = Line(H_4, H_5)
H_7: O_3 + (1, 1)
H_8: O_3 + (5, 5)
G = (12.12, -7.34)
θ = 0 rad (slider from 0 to 45*°, step 0.1*°)
pic50 = Rotate(pic49, θ, G)
t = 0 (slider from 0 to 1, step 0.01)
I: G + (9, 0)
pic51 = Translate(pic50, Vector[(t * Vector[G, H_7 - (1.5, 1.5)])])
M_{52} = Line(H_7, H_8)
H_9: O_3 + (3, 3)
text47: "我們任取一點A(1,1)，\;" + (LaTeX[If[s ≥ 48, "再取一點B(5,5)"]]) + "\\" + (LaTeX[If[s ≥ 49, "將直線\overleftrightarrow{AB}連起來"]]) + "" + (LaTeX[If[s ≥ 53, "再選擇直線上一點C(3,3)"]]) + ""
J = Point(u_{x3})
text55: "坐標平面上，\;若直線L是水平線，\;則規定L的斜角是0°；\;若L不是水平線，\;也不是鉛直線，\;則L與x軸\\
相交，\;且L的斜率m不為0，\;我們就m值討論如下：\\" + (LaTeX[If[s ≥ 56, "(1)\;若m>0時，\;L與x軸正向有一夾角α，\;0°<α<90°；"]]) + "\\" + (LaTeX[If[s ≥ 57, "(2)\;若m<0時，\;L與x軸正向有一夾角α，\;90°<α<180°；"]]) + "\\" + (LaTeX[If[s ≥ 58, "此時α稱為直線L的斜角，\;鉛直線的斜角定義為90°；\;由斜率與正切(tan)的意義可知：\;m=tanθ"]]) + ""
α = 0.9599310885968813 rad (slider from 0 to 180*°, step 1*°)
L_J: y - y(J) = (tan(α) * (x - x(J)))
β = Angle(J + (1, 0), J, J + (cos(α), sin(α)))
m_4: tan(α)
text56: "" + (LaTeX[If[m_4 > 0, "m>0", If[m_4  <  0, "m<0", "m=0"]]]) + ""
text58: "m=tan" + (LaTeX[β]) + "=" + (LaTeX[tan(β)]) + "" + (LaTeX[If[m_4 > 0, ">0", If[m_4  <  0, "<0"]]]) + ""
L_1 = Line(E_1, E_2)
text12: "觀念整理：\;\\" + (LaTeX[If[s ≥ 13, "(1)\;當直線由左下到右上傾斜時，\;其斜率為正。"]]) + "\\" + (LaTeX[If[s ≥ 14, "(2)\;當直線由左上到右下傾斜時，\;其斜率為負。"]]) + "\\" + (LaTeX[If[s ≥ 15, "(3)\;當直線成水平時，\;其斜率為0。"]]) + "\\" + (LaTeX[If[s ≥ 16, "(4)\;當直線成鉛直時，\;斜率不存在。"]]) + ""
text5: "定義：\;直線斜率\\
設A(x_1,y_1),\;B(x_2,y_2)為直線L上的相異兩點，\;則：\\" + (LaTeX[If[s ≥ 6, "(1)\;若L不垂直x軸，\;即x_1≠x_2，\;則L的斜率為m=\frac{y_2-y_1}{x_2-x_1}"]]) + "\\" + (LaTeX[If[s ≟ 7, "(2)\;若L垂直x軸，\;即x_1=x_2，\;則定義L的斜率不存在。"]]) + ""
text9: "" + (LaTeX[If[s ≟ 9, "該直線的斜率為：\\m=\frac{4-1}{7-3}=\frac{3}{4}"]]) + "" + (LaTeX[If[s ≟ 11, "該直線的斜率為：\\m=\frac{-1-5}{2-(-1)}=-2"]]) + ""
text19: "∵0<m_1<m_2,\;" + (LaTeX[If[s ≥ 20, "\\且0>m_4>m_3"]]) + "\\" + (LaTeX[If[s ≥ 21, "∴m_3<m_4<m_1<m_2"]]) + "\\" + (LaTeX[If[s ≥ 22, "故選(C)"]]) + ""
M_b: y - y(F) = (1.3 * (x - x(F)))