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45 lines
1.1 KiB
45 lines
1.1 KiB
2 years ago
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根据直角顶点的不同，应该有三种可能：
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- $A$ 线
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$\triangle ABH_1 \sim AOP$
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$4:(1-(-1))=1:OP_1$
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$OP_1=\frac{1}{2}$
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$\therefore P_1(0,-\frac{1}{2})$
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因为直角三角形是平行四边形，可以使用对角线的相关定理，$Q_x-1=1+0$
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$\therefore Q_x=2$
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$4+(-\frac{1}{2})=0+Q_y$
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$\therefore Q_Y=\frac{7}{2}$
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- $B$ 线
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同理，$\triangle P_2H_2B \sim \triangle ABH_1$
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根据比例关系，$P_2(0,\frac{9}{2})$
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$$
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\large \left\{\begin{matrix}
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0-1=1+x & \\
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\frac{9}{2}=4+y &
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\end{matrix}\right. $$
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解得:
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$x=-2,y=\frac{1}{2}$
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- $C$ 圆
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先计算$A$B的距离=$\sqrt{4+16}=2\sqrt{5}$
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半径就是$\sqrt{5}$
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直线方程$y=kx+b,A(-1,0),B(1,4)$
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$$
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\large \left\{\begin{matrix}
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-k+b=0 & \\
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k+b=4 &
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\end{matrix}\right. $$
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$\therefore b=2,k=2$
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方程$y=2x+2$
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$N$点坐标可求：$N(0,2)$
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$\therefore P_3(0,2+\sqrt{5}),P_4(0,2-\sqrt{5})$
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此时发现，正好这两个点是矩形的另外两个坐标点，即$Q_1(0,2+\sqrt{5}),Q_2(0,2-\sqrt{5})$
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