问题描述:
matlab二元函数
请问各路大神,MatLab中我求二元二次函数,怎么会这样的结果,看不懂啊,
>> [xc,yc]=solve('2*(xa-xc)^2+2*(ya-yc)^2=(xb-xa)^2+(yb-ya)^2','2*(xb-xc)^2+2*(yb-yc)^2=(xb-xa)^2+(yb-ya)^2')
xc =
xc-yb+yc
xc+yb-yc
-i*((xc+i*(yb-yc))*xc-xc^2+yb*yc-yc^2-yb*ya+ya*yc)/(yb-yc)
i*((xc-i*(yb-yc))*xc-xc^2+yb*yc-yc^2-yb*ya+ya*yc)/(yb-yc)
yc =
ya+xc-yc
-ya+yc+xc
xc+i*(yb-yc)
xc-i*(yb-yc)
请问各路大神,MatLab中我求二元二次函数,怎么会这样的结果,看不懂啊,
>> [xc,yc]=solve('2*(xa-xc)^2+2*(ya-yc)^2=(xb-xa)^2+(yb-ya)^2','2*(xb-xc)^2+2*(yb-yc)^2=(xb-xa)^2+(yb-ya)^2')
xc =
xc-yb+yc
xc+yb-yc
-i*((xc+i*(yb-yc))*xc-xc^2+yb*yc-yc^2-yb*ya+ya*yc)/(yb-yc)
i*((xc-i*(yb-yc))*xc-xc^2+yb*yc-yc^2-yb*ya+ya*yc)/(yb-yc)
yc =
ya+xc-yc
-ya+yc+xc
xc+i*(yb-yc)
xc-i*(yb-yc)
问题解答:
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