已知两点 a b 坐标,cab为90度,ac为L求 C点做坐标,C点坐标貌似在A点左边右边好像都行,求出答案格式为 x=公式1 y=公式2 公式中放已知常量 x1 y1 x2 y2 L. 已知答案留下 联系方式,三个月会员你们懂的,谢谢
设:ab所在直线方程y=kx+b
将(x1,y1),(x2,y2)代入
y1=kx1+b
y2=kx2+b
联立求解得:
因为ac⊥ab
所以ac斜率为:
ac所在直线方程y=(x2-x1)/(y2-y1)x+c
将(x1,y1)代入,求出c
ac所在直线方程y=(x2-x1)/(y2-y1)x+y1-x1(x2-x1)/(y1-y2)
又ac^2=(x-x1)^2+(y-y1)^2=x^2-2xx1+x1^2+[(x2-x1)/(y2-y1)x+y1-x1(x2-x1)/(y1-y2)-y1]=L^2
整理得:x^2-[2x1-(x2-x1)/(y2-y1)]x+[x1^2--x1(x2-x1)/(y1-y2)-L^2=0
△=[(x2-x1)/(y2-y1)] ^2-4[x1^2-x1(x2-x1)/(y1-y2)-L^2]
x=-b±√△/2a
解得:
x=[2x1-(x2-x1)/(y2-y1)]/2+√{[(x2-x1)/(y2-y1)] ^2-4[x1^2-x1(x2-x1)/(y1-y2)-L^2]}/2
或
x=[2x1-(x2-x1)/(y2-y1)]/2-√{[(x2-x1)/(y2-y1)] ^2-4[x1^2-x1(x2-x1)/(y1-y2)-L^2]}/2
(C在a上方,所以舍去)
答案:
x=[2x1-(x2-x1)/(y2-y1)]/2+√{[(x2-x1)/(y2-y1)] ^2-4[x1^2-x1(x2-x1)/(y1-y2)-L^2]}/2
y=(x2-x1)/(y2-y1)[2x1-(x2-x1)/(y2-y1)]/2+(x2-x1)/(y2-y1)*√{[(x2-x1)/(y2-y1)] ^2-4[x1^2-x1(x2-x1)/(y1-y2)-L^2]}/2+y1-x1(x2-x1)/(y1-y2)