Re: [中學] 截距式與算幾

※ 引述《choun (原來跑步這麼舒服)》之銘言： : https://i.imgur.com/a/zQEDjTb : 如圖~~ : 感覺這題應該是很常見的難題~ : 但是沒有查到、然後怎麼拼算幾… 就是拼不出一樣的答案… : 還請大神們幫幫忙！！ : 謝謝。 y - 2 = -M(x - 3) => OA = (3M + 2)/M OB = 3M + 2 M = OB/OA => OA * OB = 3OB + 2OA AB = [(3M + 2)/M]√[1 + M^2] => PA/AB = 3M/(3M + 2) PB/AB = 2/(3M + 2) => 1/PA + 1/PB = (1/6)(3M + 2)/√[1 + M^2] = (1/6)(3OB + 2OA)/√[OA^2 + OB^2] = (1/6)(OA * OB)/√[OA^2 + OB^2] <= (1/6)OP = (1/6)√13 等號成立在M = 3/2 即直線:y - 2 = -(3/2)(x - 3) -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 117.56.175.175 (臺灣) ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1669851531.A.D8D.html