已知正方形ABCD,点A(-2,-3),C(1,1)
求顶点D的坐标 D(-2,1),B(1,-3)能不能指点一下思路，拜托了参考图，AC=√[(1-(-2))^2+(1-(-3))^2]=5，对角线交点 o(-1/2,-1)设D(x,y) 则有 (y-(-2))/(x-(-1/2))=-(-1/2-(-2))/(-1-(-3))(y+2)/(x+1/2)=-3/4-------------------------------①DO=AC/2=5/2 (x-(-1/2))^2+(y-(-1))^2=(5/2)^2 ----------------------------- ②解 ①② 得 x=-5/2,y=1/2D(-5/2,1/2)