作者Honor1984 (奈何上天造化弄人？)
看板Math
标题Re: [中学] 截距式与算几
时间Thu Dec 1 07:38:49 2022
※ 引述《choun (原来跑步这麽舒服)》之铭言：
: https://imgur.com/a/zQEDjTb
: 如图~~
: 感觉这题应该是很常见的难题~
: 但是没有查到、然後怎麽拼算几… 就是拼不出一样的答案…
: 还请大神们帮帮忙！！
: 谢谢。
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
等号成立在M = 3/2
即直线:y - 2 = -(3/2)(x - 3)
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