某学生在平面直角坐标系中绘制了一个四边形ABCD,其中A(0, 0),B(4, 0),C(4, 3),D(0, 5)。若将该四边形绕原点逆时针旋转90°,得到新四边形A'B'C'D',则点C'的坐标为___。
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利用勾股定理逆定理,设C(x,y),由AC² + BC² = AB²列方程,结合x,y为正整数且在第一象限,枚举验证可得4组解。
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