基本介紹
定理1
![定比分點](/img/2/518/wZwpmL2MzMxkTOyUTMxMzM1UTM1QDN5MjM5ADMwAjMwUzL1EzL0IzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![定比分點](/img/1/b32/wZwpmL1IDM1ADNyMTMzEDN0UTMyITNykTO0EDMwAjMwUzLzEzL1IzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![定比分點](/img/1/434/wZwpmL1gDMzgzN0MDO2UzM1UTM1QDN5MjM5ADMwAjMwUzLzgzLwIzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![定比分點](/img/2/518/wZwpmL2MzMxkTOyUTMxMzM1UTM1QDN5MjM5ADMwAjMwUzL1EzL0IzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![定比分點](/img/3/2f8/wZwpmLwIDNxAjMxMTMzEDN0UTMyITNykTO0EDMwAjMwUzLzEzLyIzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
設坐標軸上有向線段 的起點A和終點B的坐標分別為 和 分點M分 的比為 ,那么,分點M的坐標
![定比分點](/img/f/585/wZwpmLzcjNwgDO4QDO2UzM1UTM1QDN5MjM5ADMwAjMwUzL0gzLzgzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
證明: 分點M的坐標為x,那么由定理1 知
![定比分點](/img/7/8b3/wZwpmL2QjMwUTN1QTO2UzM1UTM1QDN5MjM5ADMwAjMwUzL0kzL2UzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
由此得
![定比分點](/img/f/585/wZwpmLzcjNwgDO4QDO2UzM1UTM1QDN5MjM5ADMwAjMwUzL0gzLzgzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
推論
![定比分點](/img/1/b32/wZwpmL1IDM1ADNyMTMzEDN0UTMyITNykTO0EDMwAjMwUzLzEzL1IzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![定比分點](/img/1/434/wZwpmL1gDMzgzN0MDO2UzM1UTM1QDN5MjM5ADMwAjMwUzLzgzLwIzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
設坐標軸上線段AB的端點A和B的坐標分別為, 和那么線段AB的中點的坐標
![定比分點](/img/a/554/wZwpmL2gzM0AjN0kTN2UzM1UTM1QDN5MjM5ADMwAjMwUzL5UzLxYzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
例題解析
![定比分點](/img/3/2f8/wZwpmLwIDNxAjMxMTMzEDN0UTMyITNykTO0EDMwAjMwUzLzEzLyIzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![定比分點](/img/3/2f8/wZwpmLwIDNxAjMxMTMzEDN0UTMyITNykTO0EDMwAjMwUzLzEzLyIzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
【 例1】 已知有兩點P(3,-2),P(-9,4),線段PP與x軸的交點P分有向線段PP所成比為 ,則有 是多少?並求P點橫坐標。
![定比分點](/img/b/2e0/wZwpmLyQDMzQDM1kDO2UzM1UTM1QDN5MjM5ADMwAjMwUzL5gzLwUzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
![定比分點](/img/2/047/wZwpmL2IDM5kTO3ITO2UzM1UTM1QDN5MjM5ADMwAjMwUzLykzL2IzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![定比分點](/img/7/17b/wZwpmLyITM0MTNxQDO2UzM1UTM1QDN5MjM5ADMwAjMwUzL0gzL4EzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
解:設 ,則有 得
![定比分點](/img/3/2f8/wZwpmLwIDNxAjMxMTMzEDN0UTMyITNykTO0EDMwAjMwUzLzEzLyIzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
評註:先由起點、分點、終點的縱坐標求出 ,進一步再得到分點的橫坐標。
【 例2】 已知平行四邊形ABCD的三個頂點A(-1,-2),B(3,4),C(0,3),則頂點D的坐標為多少?
解:設平行四邊形ABCD的對角線AC,BD的交點為E(x,y),即E為AC的中點,所以
![定比分點](/img/4/d2d/wZwpmLzgzN4YTO1EjN2UzM1UTM1QDN5MjM5ADMwAjMwUzLxYzLzAzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
![定比分點](/img/6/69f/wZwpmLzUDOwcjNzQTO2UzM1UTM1QDN5MjM5ADMwAjMwUzL0kzL4QzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
即E點的坐標為 。
![定比分點](/img/b/ca0/wZwpmL2YTOzkzNwUzN2UzM1UTM1QDN5MjM5ADMwAjMwUzL1czL0EzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![定比分點](/img/6/a7e/wZwpmLyITO2YDN2MDO2UzM1UTM1QDN5MjM5ADMwAjMwUzLzgzL3IzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
又因為E為BD的中點,所以解得 。
評註: 利用平行四邊形性質。
【 例3】 在平面上有五個整點(坐標為整數的點),證明其中至少有兩個點的連線的中點也是整點。
證明: 設A,B,C,D,E是五個整點,則每個點的坐標的奇偶不外四種可能,就是(偶,偶)、(奇,奇)、(奇,偶)和(偶,奇)。我們取四個點A、B、C、D,它們的坐標的“最壞”情形是(偶,偶)、(奇,奇)、(奇,偶)、(偶,奇)。因為這時四個點中任意兩個點的連線的中點都不是整點,第五個點E的坐標只能是上面說的四種情形之一,但不論是哪種情形,容易驗證E與A、B、C、D中的某一點的連線的中點必是整點。
![定比分點](/img/7/3b4/wZwpmL3gzN2czM0czMxMzM1UTM1QDN5MjM5ADMwAjMwUzL3MzL3UzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![定比分點](/img/2/35d/wZwpmLyYDN0gzN1czNxMzM1UTM1QDN5MjM5ADMwAjMwUzL3czL2UzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
【 例4】 在點 和 處各放置質量為m和m的質點,求證:這兩個質點組成的質點系的重心的坐標為
![定比分點](/img/8/18e/wZwpmLxEDO2YTN2MDN3UzM1UTM1QDN5MjM5ADMwAjMwUzLzQzL2MzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
![定比分點](/img/1/02f/wZwpmL4gzM5QjN1YzN2UzM1UTM1QDN5MjM5ADMwAjMwUzL2czL4QzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![定比分點](/img/e/bb8/wZwpmL3UDM4YjN3ADM3UzM1UTM1QDN5MjM5ADMwAjMwUzLwAzL1QzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
在n個點 處各放置質量為 的質點,求證:這n個質點組成的質點系的重心的坐標為
![定比分點](/img/4/f23/wZwpmL3ETN5gTO3MDN3UzM1UTM1QDN5MjM5ADMwAjMwUzLzQzLwYzLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
證明:兩個質點組成的質點系的重心G線上段PP上,並且滿足條件
![定比分點](/img/4/05a/wZwpmL4ADO5gTMzQjN2UzM1UTM1QDN5MjM5ADMwAjMwUzL0YzLwczLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
即
![定比分點](/img/a/05b/wZwpmL1YjM0MTM5MzM3UzM1UTM1QDN5MjM5ADMwAjMwUzLzMzLwgzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
所以
![定比分點](/img/8/36d/wZwpmLwUzNyMDOzgTN2UzM1UTM1QDN5MjM5ADMwAjMwUzL4UzLwYzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
所以重心G的坐標
![定比分點](/img/e/4f9/wZwpmL4YTO0ATMxQzM3UzM1UTM1QDN5MjM5ADMwAjMwUzL0MzL4MzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
![定比分點](/img/9/ea5/wZwpmL4UDMykjM4QDO2UzM1UTM1QDN5MjM5ADMwAjMwUzL0gzLxIzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
一般情形請讀者用數學歸納法證明。
![定比分點](/img/1/02f/wZwpmL4gzM5QjN1YzN2UzM1UTM1QDN5MjM5ADMwAjMwUzL2czL4QzLt92YucmbvRWdo5Cd0FmL0E2LvoDc0RHa.jpg)
![定比分點](/img/2/396/wZwpmLygjM1YjN0QzM3UzM1UTM1QDN5MjM5ADMwAjMwUzL0MzL3UzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![定比分點](/img/2/396/wZwpmLygjM1YjN0QzM3UzM1UTM1QDN5MjM5ADMwAjMwUzL0MzL3UzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![定比分點](/img/9/6b6/wZwpmL0QzM2kzN2czN2UzM1UTM1QDN5MjM5ADMwAjMwUzL3czL1gzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![定比分點](/img/9/6b6/wZwpmL0QzM2kzN2czN2UzM1UTM1QDN5MjM5ADMwAjMwUzL3czL1gzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![定比分點](/img/e/f2e/wZwpmLwcDO2ETO2czN2UzM1UTM1QDN5MjM5ADMwAjMwUzL3czL3gzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
![定比分點](/img/e/f2e/wZwpmLwcDO2ETO2czN2UzM1UTM1QDN5MjM5ADMwAjMwUzL3czL3gzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
![定比分點](/img/4/e3d/wZwpmL2gTN4kTM4UzN2UzM1UTM1QDN5MjM5ADMwAjMwUzL1czL4AzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
![定比分點](/img/4/e3d/wZwpmL2gTN4kTM4UzN2UzM1UTM1QDN5MjM5ADMwAjMwUzL1czL4AzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
【 例5】已知n個點 ,在有向線段 上取一點G,使G分 的比為1:1;在有向線段 上取一點G,使G分 的比為1:2;在有向線段 上取一點G使G分 的比為1:3;......;在有向線段 上取一點G,使G分 的比為1:n-1,求證:最後的分點G的坐標為
![定比分點](/img/2/08f/wZwpmL2UDO3cTO0QDO2UzM1UTM1QDN5MjM5ADMwAjMwUzL0gzL2YzLt92YucmbvRWdo5Cd0FmLyE2LvoDc0RHa.jpg)
點G叫作已知的n個點P,P,…,P的(幾何)重心(圖1)。
![圖1](/img/2/33d/wZwpmL3UzM0QTM5gDN3UzM1UTM1QDN5MjM5ADMwAjMwUzL4QzLyAzLt92YucmbvRWdo5Cd0FmLwE2LvoDc0RHa.jpg)
![定比分點](/img/2/86c/wZwpmL2gDO2ITOzEDNzEzM1UTM1QDN5MjM5ADMwAjMwUzLxQzLyUzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
特別地,以 為頂點的三角形的(幾何)重心的坐標為
![定比分點](/img/f/571/wZwpmLxYTM2UzM5YTO2UzM1UTM1QDN5MjM5ADMwAjMwUzL2kzLyAzLt92YucmbvRWdo5Cd0FmLzE2LvoDc0RHa.jpg)
證明: 設例4中的n個質點的質量都相等,這時n個質點的力學重心即是n個點P,P,…,P的幾何重心G,所以G的坐標為
![定比分點](/img/a/6f1/wZwpmLxYDOzYTMwUzN2UzM1UTM1QDN5MjM5ADMwAjMwUzL1czLxczLt92YucmbvRWdo5Cd0FmLxE2LvoDc0RHa.jpg)
不利用例4也可獨立證明。