在直角坐標平面上一條連接(2x-1,2x+1)和(x-6,1-3x)兩點的線段長度是17。求x的可能性?
回答 (2)
2017-10-02 10:03 am
[(2x-1)-(x-6)]^2+[(2x+1)-(1-3x)]^2=17^2
(x+5)^2+(5x)^2=289
x^2+10x+25+25x^2=289
26x^2+10x-264=0
x=3 or x=-44/13
(x+5)^2+(5x)^2=289
x^2+10x+25+25x^2=289
26x^2+10x-264=0
x=3 or x=-44/13
2017-10-01 1:48 pm
17^2=[(2x-1)-(x-6)]^2+[(2x+1)-(1-3x)]^2
289=(x+5)^2+(5x)^2
289=26x^2+10x+25
26x^2+10x-264=0
13x^2+5x-132=0
(13x+44)(x-3)=0
x= - 44/13 or 3
289=(x+5)^2+(5x)^2
289=26x^2+10x+25
26x^2+10x-264=0
13x^2+5x-132=0
(13x+44)(x-3)=0
x= - 44/13 or 3
收錄日期: 2021-04-18 17:50:35
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171001024732AAyXE0l