如图,正方形ABCD与正方形EFGH是位似图形,已知A(0,5),D(0,3),E(0,1),H(0,4),则位似中心的坐标可能是.
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没有指明对应点时,答案不唯一,注意分类讨论.
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位似图形对应点的连线交点为位似中心. 当正方形ABCD∽正方形 EFGH时,A与E,B与F为对应点,直线AE为x=0,设直线BF的解析式为$y = k x + b ( k \neq 0 )$,则$\left\{\begin{array} {c} {- 2 k + b = 5},\\ {3 k + b = 1},\end{array} \right.$解得$\left\{\begin{array} {l} {k = - \frac {4} {5}},\\ {b = \frac {17} {5}},\end{array} \right.$故直线BF的解析式为$y = - \frac {4} {5} x + \frac {17} {5}$,与x=0联立解得$y = \frac {17} {5}$,即位似中心是$( 0,\frac {17} {5} )$;当正方形ABCD∽正方形GHEF时,C与E是对应点,设直线CE的解析式为$y = a x + c ( a \neq 0 )$,则$\left\{\begin{array} {l} {- 2 a + c = 3},\\ {c = 1},\end{array} \right.$解得$\left\{\begin{array} {l} {a = - 1},\\ {c = 1},\end{array} \right.$故直线CE的解析式为$y = - x + 1$. D与F是对应点,设直线DF的解析式为$y = d x + e ( d \neq 0 )$,则$\left\{\begin{array} {l} {3 d + e = 1},\\ {e = 3},\end{array} \right.$,解得$\left\{\begin{array} {l} {d = - \frac {2} {3}},\\ {e = 3},\end{array} \right.$故直线DF的解析式为$y = - \frac {2} {3} x + 3$,联立直线CE,DF的解析式得$\left\{\begin{array} {l} {y = - \frac {2} {3} x + 3},\\ {y = - x + 1},\end{array} \right.$解得$\left\{\begin{array} {l} {x = - 6},\\ {y = 7},\end{array} \right.$,即位似中心是$( - 6,7 )$;
当正方形ABCD∽正方形HEFG,或当正方形ABCD∽正方形FGHE时,对应点连线不交于一点,不符合题意.
综上所述,所求位似中心的坐标为$( 0,\frac {17} {5} )$或$( - 6,7 )$.