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Title: [筆記] 仿射變換性質(zhì)的代數(shù)證明
文章目錄
- I. 仿射變換的代數(shù)表示
- II. 仿射變換的性質(zhì)
- III. 同素性的代數(shù)證明
- 1. 點(diǎn)變換為點(diǎn)
- 2. 直線變換為直線
- IV. 結(jié)合性的代數(shù)證明
- 1. 直線上一點(diǎn)映射為直線上一點(diǎn)
- 2. 直線外一點(diǎn)映射為直線外一點(diǎn)
- V. 保持單比的代數(shù)證明
- VI. 平行性的代數(shù)證明
- 參考文獻(xiàn)
I. 仿射變換的代數(shù)表示
平面上點(diǎn) P ( x , y ) P(x,y) P(x,y) 經(jīng)過仿射變換 T T T 變?yōu)辄c(diǎn) P ′ ( x ′ , y ′ ) P'(x', y') P′(x′,y′), 則兩點(diǎn)坐標(biāo) ( x , y ) (x,y ) (x,y) 和 ( x ′ , y ′ ) (x', y') (x′,y′) 之間的關(guān)系即為仿射變換的代數(shù)表示. 需注意仿射坐標(biāo)系不一定是直角坐標(biāo)系.
平面上的仿射變換在仿射坐標(biāo)系下的代數(shù)表示為
{ x ′ = a 11 x + a 12 y + a 13 y ′ = a 21 x + a 22 y + a 23 (I-1) \left\{ \begin{aligned} x' = a_{11} x + a_{12} y + a_{13}\\ y' = a_{21} x + a_{22} y + a_{23} \end{aligned} \right. \tag{I-1} {x′=a11?x+a12?y+a13?y′=a21?x+a22?y+a23??(I-1)
其中
Δ = ∣ a 11 a 12 a 21 a 22 ∣ ≠ 0 (I-2) \Delta = \left|\begin{matrix}a_{11} &a_{12} \\ a_{21} & a_{22}\end{matrix}\right| \neq 0 \tag{I-2} Δ= ?a11?a21??a12?a22?? ?=0(I-2)
也就是仿射變換是可逆變換, 其逆變換可以寫成
{ x = a 11 ′ x ′ + a 12 ′ y ′ + a 13 ′ y = a 21 ′ x ′ + a 22 ′ y ′ + a 23 ′ (I-3) \left\{ \begin{aligned} x = a'_{11} x' + a'_{12} y' + a'_{13}\\ y = a'_{21} x' + a'_{22} y' + a'_{23} \end{aligned} \right. \tag{I-3} {x=a11′?x′+a12′?y′+a13′?y=a21′?x′+a22′?y′+a23′??(I-3)
其中
Δ ′ = ∣ a 11 ′ a 12 ′ a 21 ′ a 22 ′ ∣ ≠ 0 (I-4) \Delta' = \left|\begin{matrix}a'_{11} &a'_{12} \\ a'_{21} & a'_{22}\end{matrix}\right| \neq 0 \tag{I-4} Δ′= ?a11′?a21′??a12′?a22′?? ?=0(I-4)
可知
[ a 11 a 12 a 21 a 22 ] [ a 11 ′ a 12 ′ a 21 ′ a 22 ′ ] = [ 1 0 0 1 ] (I-5) \begin{bmatrix}a_{11} &a_{12} \\ a_{21} & a_{22}\end{bmatrix} \begin{bmatrix}a'_{11} &a'_{12} \\ a'_{21} & a'_{22}\end{bmatrix} = \begin{bmatrix}1 &0 \\ 0 & 1 \end{bmatrix} \tag{I-5} [a11?a21??a12?a22??][a11′?a21′??a12′?a22′??]=[10?01?](I-5)
及
[ a 11 ′ a 12 ′ a 21 ′ a 22 ′ ] [ a 11 a 12 a 21 a 22 ] = [ 1 0 0 1 ] (I-6) \begin{bmatrix}a'_{11} &a'_{12} \\ a'_{21} & a'_{22}\end{bmatrix} \begin{bmatrix}a_{11} &a_{12} \\ a_{21} & a_{22}\end{bmatrix} = \begin{bmatrix}1 &0 \\ 0 & 1 \end{bmatrix} \tag{I-6} [a11′?a21′??a12′?a22′??][a11?a21??a12?a22??]=[10?01?](I-6)
II. 仿射變換的性質(zhì)
- 保持同素性 (點(diǎn)變換為點(diǎn), 直線變換為直線)
- 保持結(jié)合性 (點(diǎn)和直線的結(jié)合關(guān)系)
- 保持共線三點(diǎn)的單比不變
- 保持直線的平行性
III. 同素性的代數(shù)證明
1. 點(diǎn)變換為點(diǎn)
定義式 (I-1) 即是證明.
2. 直線變換為直線
假設(shè)有一直線方程為
a x + b y + c = 0 (III-2-1) ax + by + c =0 \tag{III-2-1} ax+by+c=0(III-2-1)
其中 a 2 + b 2 ≠ 0 a^2 + b^2 \neq 0 a2+b2=0.
在仿射變換下有式 (I-3), 代入式 (III-2-1) 得到
a ( a 11 ′ x ′ + a 12 ′ y ′ + a 13 ′ ) + b ( a 21 ′ x ′ + a 22 ′ y ′ + a 23 ′ ) + c = 0 ? ( a a 11 ′ + b a 21 ′ ) x ′ + ( a a 12 ′ + b a 22 ′ ) y ′ + ( a a 13 ′ + b a 23 ′ + c ) = 0 (III-2-2) \begin{aligned} a(a'_{11} x' + a'_{12} y' + a'_{13}) + b(a'_{21} x' + a'_{22} y' + a'_{23}) + c =0\\ \Rightarrow \quad (a a'_{11}+ b a'_{21}) x' + (aa'_{12} + b a'_{22}) y' +(a a'_{13} + b a'_{23} +c) = 0 \end{aligned} \tag{III-2-2} a(a11′?x′+a12′?y′+a13′?)+b(a21′?x′+a22′?y′+a23′?)+c=0?(aa11′?+ba21′?)x′+(aa12′?+ba22′?)y′+(aa13′?+ba23′?+c)=0?(III-2-2)
另外, 由仿射變換的可逆性質(zhì)式 (I-3) 可知
∣ a 11 ′ a 21 ′ a 12 ′ a 22 ′ ∣ ≠ 0 (III-2-3) \left|\begin{matrix} a'_{11} &a'_{21} \\ a'_{12} &a'_{22} \end{matrix} \right| \neq 0 \tag{III-2-3} ?a11′?a12′??a21′?a22′?? ?=0(III-2-3)
則使得下式
[ a a 11 ′ + b a 21 ′ a a 12 ′ + b a 22 ′ ] = [ a 11 ′ a 21 ′ a 12 ′ a 22 ′ ] [ a b ] = [ 0 0 ] (III-2-4) \begin{bmatrix}a a'_{11}+ b a'_{21} \\ aa'_{12} + b a'_{22} \end{bmatrix}=\begin{bmatrix} a'_{11} &a'_{21} \\ a'_{12} &a'_{22} \end{bmatrix} \begin{bmatrix}a\\b \end{bmatrix} = \begin{bmatrix}0\\0 \end{bmatrix} \tag{III-2-4} [aa11′?+ba21′?aa12′?+ba22′??]=[a11′?a12′??a21′?a22′??][ab?]=[00?](III-2-4)
成立的解僅為 ( a , b ) = ( 0 , 0 ) (a, b) = (0, 0) (a,b)=(0,0), 但與條件矛盾. 故 ( a a 11 ′ + b a 21 ′ ) (a a'_{11}+ b a'_{21}) (aa11′?+ba21′?) 與 ( a a 12 ′ + b a 22 ′ ) (aa'_{12} + b a'_{22}) (aa12′?+ba22′?) 不能同時(shí)為零.
即式 (III-2-2) 中變量 x ′ x' x′ 和 y ′ y' y′ 前的系數(shù)滿足
( a a 11 ′ + b a 21 ′ ) 2 + ( a a 12 ′ + b a 22 ′ ) 2 ≠ 0 (III-2-5) (a a'_{11}+ b a'_{21})^2 + (aa'_{12} + b a'_{22})^2 \neq 0 \tag{III-2-5} (aa11′?+ba21′?)2+(aa12′?+ba22′?)2=0(III-2-5)
即式 (III-2-2) 必然為直線方程. 所以直線方程經(jīng)過仿射變換后仍然為直線方程, 直線經(jīng)過仿射變換后仍為直線.
IV. 結(jié)合性的代數(shù)證明
1. 直線上一點(diǎn)映射為直線上一點(diǎn)
已知一點(diǎn) ( x 1 , y 1 ) (x_1, y_1) (x1?,y1?) 在直線 a x + b y + c = 0 ax+by+c =0 ax+by+c=0 上 ? \Longleftrightarrow ? 滿足 a x 1 + b y 1 + c = 0 ax_1 +by_1 +c =0 ax1?+by1?+c=0.
下面證明仿射變換后的新點(diǎn)在仿射變換后的新直線上.
點(diǎn) ( x 1 , y 1 ) (x_1, y_1) (x1?,y1?) 經(jīng)過仿射變換后得到新點(diǎn) ( x 1 ′ , y 1 ′ ) (x'_1, y'_1) (x1′?,y1′?)
{ x 1 ′ = a 11 x 1 + a 12 y 1 + a 13 y 1 ′ = a 21 x 1 + a 22 y 1 + a 23 (IV-1-1) \left\{ \begin{aligned} x'_1 = a_{11} x_1 + a_{12} y_1 + a_{13}\\ y'_1 = a_{21} x_1 + a_{22} y_1 + a_{23} \end{aligned} \right. \tag{IV-1-1} {x1′?=a11?x1?+a12?y1?+a13?y1′?=a21?x1?+a22?y1?+a23??(IV-1-1)
由同素性式 (III-2-2), 直線 a x + b y + c = 0 ax+by+c =0 ax+by+c=0 經(jīng)過仿射變換后得到新的直線方程
( a a 11 ′ + b a 21 ′ ) x ′ + ( a a 12 ′ + b a 22 ′ ) y ′ + ( a a 13 ′ + b a 23 ′ + c ) = 0 (IV-1-2) (a a'_{11}+ b a'_{21}) x' + (aa'_{12} + b a'_{22}) y' +(a a'_{13} + b a'_{23} +c) = 0 \tag{IV-1-2} (aa11′?+ba21′?)x′+(aa12′?+ba22′?)y′+(aa13′?+ba23′?+c)=0(IV-1-2)
將仿射變換后的新點(diǎn)坐標(biāo)式 (IV-1-1) 代入上式的左側(cè)得到
LHS? = ( a a 11 ′ + b a 21 ′ ) ( a 11 x 1 + a 12 y 1 + a 13 ) ￣ + ( a a 12 ′ + b a 22 ′ ) ( a 21 x 1 + a 22 y 1 + a 23 ) ￣ + ( a a 13 ′ + b a 23 ′ + c ) = ( a a 11 ′ + b a 21 ′ ) a 11 x 1 + ( a a 12 ′ + b a 22 ′ ) a 21 x 1 + ( a a 11 ′ + b a 21 ′ ) a 12 y 1 + ( a a 12 ′ + b a 22 ′ ) a 22 y 1 + ( a a 11 ′ + b a 21 ′ ) a 13 + ( a a 12 ′ + b a 22 ′ ) a 23 + ( a a 13 ′ + b a 23 ′ + c ) = ( a a 11 ′ a 11 + a a 12 ′ a 21 + b a 21 ′ a 11 + b a 22 ′ a 21 ) x 1 + ( a a 11 ′ a 12 + a a 12 ′ a 22 + b a 21 ′ a 12 + b a 22 ′ a 22 ) y 1 + ( a a 11 ′ a 13 + a a 12 ′ a 23 + a a 13 ′ + b a 21 ′ a 13 + b a 22 ′ a 23 + b a 23 ′ ) + c = [ a b ] [ a 11 ′ a 12 ′ a 21 ′ a 22 ′ ] [ a 11 a 21 ] x 1 + [ a b ] [ a 11 ′ a 12 ′ a 21 ′ a 22 ′ ] [ a 12 a 22 ] y 1 + [ a b ] { [ a 11 ′ a 12 ′ a 21 ′ a 22 ′ ] [ a 13 a 23 ] + [ a 13 ′ a 23 ′ ] } + c (IV-1-3) \begin{aligned} \text{LHS}\ =\ &(a a'_{11}+ b a'_{21})\underline{(a_{11} x_1 + a_{12} y_1 + a_{13})} + (aa'_{12} + b a'_{22}) \underline{(a_{21} x_1 + a_{22} y_1 + a_{23})} + (a a'_{13} + b a'_{23} +c)\\ =\ & (a a'_{11}+ b a'_{21})a_{11} x_1 + (aa'_{12} + b a'_{22}) a_{21} x_1 \\ {+} & (a a'_{11}+ b a'_{21}) a_{12} y_1 +(aa'_{12} + b a'_{22}) a_{22} y_1 \\ {+}& (a a'_{11}+ b a'_{21}) a_{13} + (aa'_{12} + b a'_{22}) a_{23} + (a a'_{13} + b a'_{23} +c)\\ =\ & (a a'_{11}a_{11}+ aa'_{12} a_{21} + b a'_{21}a_{11} + b a'_{22} a_{21}) x_1 \\ {+}& (a a'_{11}a_{12}+ a a'_{12} a_{22} + b a'_{21} a_{12} + b a'_{22} a_{22}) y_1\\ {+}& (a a'_{11} a_{13}+ aa'_{12} a_{23} + a a'_{13} + b a'_{21} a_{13} + b a'_{22} a_{23} + b a'_{23}) \\ +& c\\ =\ & \begin{bmatrix} a &b\end{bmatrix} \begin{bmatrix} a'_{11} &a'_{12} \\ a'_{21} & a'_{22}\end{bmatrix} \begin{bmatrix} a_{11} \\ a_{21}\end{bmatrix} x_1 \\ {+}& \begin{bmatrix} a &b\end{bmatrix} \begin{bmatrix} a'_{11} &a'_{12} \\ a'_{21} & a'_{22}\end{bmatrix} \begin{bmatrix} a_{12} \\ a_{22}\end{bmatrix} y_1 \\ {+}& \begin{bmatrix} a &b\end{bmatrix} \left\{\begin{bmatrix} a'_{11} &a'_{12} \\ a'_{21} & a'_{22}\end{bmatrix} \begin{bmatrix} a_{13} \\ a_{23}\end{bmatrix}+ \begin{bmatrix} a'_{13} \\ a'_{23}\end{bmatrix}\right\} \\ {+}& c \end{aligned} \tag{IV-1-3} LHS?=?=?++=?+++=?+++?(aa11′?+ba21′?)(a11?x1?+a12?y1?+a13?)?+(aa12′?+ba22′?)(a21?x1?+a22?y1?+a23?)?+(aa13′?+ba23′?+c)(aa11′?+ba21′?)a11?x1?+(aa12′?+ba22′?)a21?x1?(aa11′?+ba21′?)a12?y1?+(aa12′?+ba22′?)a22?y1?(aa11′?+ba21′?)a13?+(aa12′?+ba22′?)a23?+(aa13′?+ba23′?+c)(aa11′?a11?+aa12′?a21?+ba21′?a11?+ba22′?a21?)x1?(aa11′?a12?+aa12′?a22?+ba21′?a12?+ba22′?a22?)y1?(aa11′?a13?+aa12′?a23?+aa13′?+ba21′?a13?+ba22′?a23?+ba23′?)c[a?b?][a11′?a21′??a12′?a22′??][a11?a21??]x1?[a?b?][a11′?a21′??a12′?a22′??][a12?a22??]y1?[a?b?]{[a11′?a21′??a12′?a22′??][a13?a23??]+[a13′?a23′??]}c?(IV-1-3)
由式 (I-6) 可知
[ a 11 ′ a 12 ′ a 21 ′ a 22 ′ ] [ a 11 a 21 ] = [ 1 0 ] (IV-1-4) \begin{bmatrix}a'_{11} &a'_{12} \\ a'_{21} & a'_{22}\end{bmatrix} \begin{bmatrix}a_{11} \\ a_{21} \end{bmatrix} = \begin{bmatrix}1 \\ 0 \end{bmatrix} \tag{IV-1-4} [a11′?a21′??a12′?a22′??][a11?a21??]=[10?](IV-1-4)
和
[ a 11 ′ a 12 ′ a 21 ′ a 22 ′ ] [ a 12 a 22 ] = [ 0 1 ] (IV-1-5) \begin{bmatrix}a'_{11} &a'_{12} \\ a'_{21} & a'_{22}\end{bmatrix} \begin{bmatrix}a_{12} \\ a_{22}\end{bmatrix} = \begin{bmatrix}0 \\ 1 \end{bmatrix} \tag{IV-1-5} [a11′?a21′??a12′?a22′??][a12?a22??]=[01?](IV-1-5)
由仿射變換式 (I-1) 可知原點(diǎn) ( 0 , 0 ) (0,0) (0,0) 的像為 ( a 13 , a 23 ) (a_{13}, a_{23}) (a13?,a23?), 即
{ x o ′ = a 11 0 + a 12 0 + a 13 = a 13 y o ′ = a 21 0 + a 22 0 + a 23 = a 23 (IV-1-6) \left\{ \begin{aligned} x'_{o} = a_{11} 0 + a_{12} 0 + a_{13} = a_{13}\\ y'_{o} = a_{21} 0 + a_{22} 0 + a_{23} = a_{23} \end{aligned} \right. \tag{IV-1-6} {xo′?=a11?0+a12?0+a13?=a13?yo′?=a21?0+a22?0+a23?=a23??(IV-1-6)
相反地, 原點(diǎn)的像 ( a 13 , a 23 ) (a_{13}, a_{23}) (a13?,a23?) 經(jīng)過仿射變換逆映射式 (I-3) 后回到原點(diǎn), 即
{ 0 = a 11 ′ a 13 + a 12 ′ a 23 + a 13 ′ 0 = a 21 ′ a 13 + a 22 ′ a 23 + a 23 ′ (IV-1-7) \left\{ \begin{aligned} 0 = a'_{11} a_{13} + a'_{12} a_{23} + a'_{13}\\ 0 = a'_{21} a_{13} + a'_{22} a_{23} + a'_{23} \end{aligned} \right. \tag{IV-1-7} {0=a11′?a13?+a12′?a23?+a13′?0=a21′?a13?+a22′?a23?+a23′??(IV-1-7)
上式寫成矩陣形式為
[ a 11 ′ a 12 ′ a 21 ′ a 22 ′ ] [ a 13 a 23 ] + [ a 13 ′ a 23 ′ ] = [ 0 0 ] (IV-1-8) \begin{bmatrix}a'_{11} & a'_{12}\\a'_{21} &a'_{22} \end{bmatrix} \begin{bmatrix}a_{13} \\ a_{23} \end{bmatrix} + \begin{bmatrix}a'_{13} \\ a'_{23} \end{bmatrix} = \begin{bmatrix}0 \\0 \end{bmatrix} \tag{IV-1-8} [a11′?a21′??a12′?a22′??][a13?a23??]+[a13′?a23′??]=[00?](IV-1-8)
將式 (IV-1-4)、(IV-1-5)、(IV-1-8) 代入式 (IV-1-3), 即仿射變換后的新點(diǎn)坐標(biāo)式 (IV-1-1) 左側(cè)為
LHS? = [ a b ] [ 1 0 ] x 1 + [ a b ] [ 0 1 ] y 1 + [ a b ] [ 0 0 ] + c = a x 1 + b y 1 + c = 0 (IV-1-9) \begin{aligned} \text{LHS}\ =\ & \begin{bmatrix} a &b\end{bmatrix} \begin{bmatrix} 1 \\ 0\end{bmatrix} x_1 {+} \begin{bmatrix} a &b\end{bmatrix} \begin{bmatrix} 0 \\ 1\end{bmatrix} y_1 {+} \begin{bmatrix} a &b\end{bmatrix} \begin{bmatrix} 0 \\ 0\end{bmatrix} {+} c\\ =\ & a x_1 {+} b y_1 {+} c\\ = \ &0 \end{aligned} \tag{IV-1-9} LHS?=?=?=??[a?b?][10?]x1?+[a?b?][01?]y1?+[a?b?][00?]+cax1?+by1?+c0?(IV-1-9)
也就是說, 點(diǎn) ( x 1 , y 1 ) (x_1, y_1) (x1?,y1?) 經(jīng)過仿射變換后得到的新點(diǎn) ( x 1 ′ , y 1 ′ ) (x'_1, y'_1) (x1′?,y1′?) 滿足直線 a x + b y + c = 0 ax+by+c =0 ax+by+c=0 進(jìn)過仿射變換后得到的新直線方程式 (IV-1-2).
即直線上一點(diǎn)經(jīng)過仿射變換后仍然在經(jīng)過仿射變換后的直線上.
2. 直線外一點(diǎn)映射為直線外一點(diǎn)
如果存在仿射變換 φ \varphi φ 可以將直線 l l l 外一點(diǎn) P P P 映射為直線 l ′ l' l′ 上一點(diǎn) P ′ P' P′, 即
φ : l ? l ′ φ : P ? P ′ P ? l , P ′ ∈ l ′ (IV-2-1) \varphi:l \mapsto l'\\ \varphi:P \mapsto P' \\ P \notin l, P' \in l'\tag{IV-2-1} φ:l?l′φ:P?P′P∈/l,P′∈l′(IV-2-1)
因?yàn)榉律渥儞Q的逆變換仍然是仿射變換, 即 φ ? 1 \varphi^{-1} φ?1 仍然是仿射變換.
由已證明的 “直線上一點(diǎn)映射為直線上一點(diǎn)”, 可知經(jīng)過逆仿射變換 φ ? 1 \varphi^{-1} φ?1, 直線 l ′ l' l′ 上點(diǎn) P ′ P' P′ 映射為直線 l l l 上點(diǎn) P P P.
即已知像點(diǎn) P ′ P' P′ 在直線 l ′ l' l′ 上時(shí), 原像點(diǎn) P P P 必在直線 l l l 上.
假設(shè)矛盾, 證畢.
V. 保持單比的代數(shù)證明
共線三點(diǎn) P ( x , y ) P(x,y) P(x,y)、 P 1 ( x 1 , y 1 ) P_1(x_1, y_1) P1?(x1?,y1?)、 P 2 ( x 2 , y 2 ) P_2(x_2, y_2) P2?(x2?,y2?) 的單比為
( P 1 P 2 P ) = x ? x 1 x ? x 2 = y ? y 1 y ? y 2 = k (V-1) (P_1 P_2 P) = \frac{x-x_1}{x - x_2}=\frac{y-y_1}{y-y_2}=k \tag{V-1} (P1?P2?P)=x?x2?x?x1??=y?y2?y?y1??=k(V-1)
即
x ? x 1 = k ( x ? x 2 ) y ? y 1 = k ( y ? y 2 ) (V-2) {x-x_1}= k(x - x_2)\\ {y-y_1}= k(y-y_2) \tag{V-2} x?x1?=k(x?x2?)y?y1?=k(y?y2?)(V-2)
經(jīng)過仿射變換后得到 P ′ ( x ′ , y ′ ) P'(x',y') P′(x′,y′)、 P 1 ′ ( x 1 ′ , y 1 ′ ) P'_1(x'_1, y'_1) P1′?(x1′?,y1′?)、 P 2 ′ ( x 2 ′ , y 2 ′ ) P'_2(x'_2, y'_2) P2′?(x2′?,y2′?). 由仿射變換的結(jié)合性可知, P ′ P' P′、 P 1 ′ P'_1 P1′?、 P 2 ′ P'_2 P2′? 三點(diǎn)仍然共線. 共線三點(diǎn)可計(jì)算單比
( P 1 ′ P 2 ′ P ′ ) = x ′ ? x 1 ′ x ′ ? x 2 ′ = y ′ ? y 1 ′ y ′ ? y 2 ′ = k ′ (V-3) (P'_1 P'_2 P') = \frac{x'-x'_1}{x' - x'_2}=\frac{y'-y'_1}{y'-y'_2}=k' \tag{V-3} (P1′?P2′?P′)=x′?x2′?x′?x1′??=y′?y2′?y′?y1′??=k′(V-3)
由仿射變換式 (I-1) 并結(jié)合式 (V-2), 可知
k ′ = x ′ ? x 1 ′ x ′ ? x 2 ′ = a 11 x + a 12 y + a 13 ? ( a 11 x 1 + a 12 y 1 + a 13 ) a 11 x + a 12 y + a 13 ? ( a 11 x 2 + a 12 y 2 + a 13 ) = a 11 ( x ? x 1 ) + a 12 ( y ? y 1 ) a 11 ( x ? x 2 ) + a 12 ( y ? y 2 ) = k a 11 ( x ? x 2 ) + k a 12 ( y ? y 2 ) a 11 ( x ? x 2 ) + a 12 ( y ? y 2 ) = k (V-4) \begin{aligned} k' &= \frac{x'-x'_1}{x' - x'_2} \\ &= \frac{a_{11} x + a_{12} y + a_{13} - (a_{11} x_1 + a_{12} y_1 + a_{13})}{a_{11} x + a_{12} y + a_{13} - (a_{11} x_2 + a_{12} y_2 + a_{13})}\\ &= \frac{a_{11} (x-x_1) + a_{12} (y-y_1)}{a_{11} (x-x_2) + a_{12} (y-y_2)}\\ &= \frac{k a_{11} (x-x_2) + k a_{12} (y-y_2)}{a_{11} (x-x_2) + a_{12} (y-y_2)}\\ &= k \end{aligned}\tag{V-4} k′?=x′?x2′?x′?x1′??=a11?x+a12?y+a13??(a11?x2?+a12?y2?+a13?)a11?x+a12?y+a13??(a11?x1?+a12?y1?+a13?)?=a11?(x?x2?)+a12?(y?y2?)a11?(x?x1?)+a12?(y?y1?)?=a11?(x?x2?)+a12?(y?y2?)ka11?(x?x2?)+ka12?(y?y2?)?=k?(V-4)
即單比保持不變.
VI. 平行性的代數(shù)證明
已知兩條直線 l 1 l_1 l1? 和 l 2 l_2 l2? 平行, 則該兩條直線的方程可寫為
{ l 1 : a x + b y + c 1 = 0 l 2 : a x + b y + c 2 = 0 (VI-1) \left\{ \begin{aligned} l_1: ax+by+c_1=0\\ l_2: ax+by+c_2=0 \end{aligned} \right. \tag{VI-1} {l1?:ax+by+c1?=0l2?:ax+by+c2?=0?(VI-1)
其中 c 1 ≠ c 2 c_1 \neq c_2 c1?=c2?. (注意仿射坐標(biāo)系不一定為直角坐標(biāo)系)
由同素性證明中式 (III-1-2) , 直線 l 1 l_1 l1? 和 l 2 l_2 l2? 經(jīng)過仿射變換的方程式分別為
l 1 ′ : ( a a 11 ′ + b a 21 ′ ) x ′ + ( a a 12 ′ + b a 22 ′ ) y ′ + ( a a 13 ′ + b a 23 ′ + c 1 ) = 0 l 2 ′ : ( a a 11 ′ + b a 21 ′ ) x ′ + ( a a 12 ′ + b a 22 ′ ) y ′ + ( a a 13 ′ + b a 23 ′ + c 2 ) = 0 l'_1: (a a'_{11}+ b a'_{21}) x' + (aa'_{12} + b a'_{22}) y' +(a a'_{13} + b a'_{23} +c_1) = 0\\ l'_2: (a a'_{11}+ b a'_{21}) x' + (aa'_{12} + b a'_{22}) y' +(a a'_{13} + b a'_{23} +c_2) = 0 l1′?:(aa11′?+ba21′?)x′+(aa12′?+ba22′?)y′+(aa13′?+ba23′?+c1?)=0l2′?:(aa11′?+ba21′?)x′+(aa12′?+ba22′?)y′+(aa13′?+ba23′?+c2?)=0
其中 a a 13 ′ + b a 23 ′ + c 1 ≠ a a 13 ′ + b a 23 ′ + c 2 a a'_{13} + b a'_{23} +c_1 \neq a a'_{13} + b a'_{23} +c_2 aa13′?+ba23′?+c1?=aa13′?+ba23′?+c2?, 而兩者的方向數(shù)相同.
可知經(jīng)過仿射變換后 l 1 ′ l'_1 l1′? 和 l 2 ′ l'_2 l2′? 仍然平行.
參考文獻(xiàn)
[1] 梅向明, 劉增賢, 王匯淳, 王智秋, 高等幾何(第四版), 高等教育出版社, 2020
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