已知正方形ABCD,点A(-2,-3),C(1,1)求顶点D的坐标

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 -------------------------------① DO=AC/2=5/2 (x-(-1/2))^2+(y-(-1))^2=(5/2)^2 ----------------------------- ② 解 ①② 得 x=-5/2,y=1/2 D(-5/2,1/2)