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求速算法：過任意兩點之二元一次方程式

Apr 21st 2014, 14:45

從那兩點看出來，當x增加2時，y減少44/21，所以每當x增加1，y就減少22/21，所以方程式應該是y=(22/21)x+什麼，再帶隨便一點，就可以得到:

y=(22/21)x+9

以上只是把想法講出來，如果只要答案用想的會很快。

參考資料： xu3

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