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第122頁

信息發(fā)布者：
??$(1)$??若??$∠PAB=90°$??
則??$P$??點(diǎn)的橫坐標(biāo)為??$-2，$??代入??$y=\frac {1} {2}x+\frac {5} {2}$??
解得：??$y=\frac {3} {2}，$??則??$P(-2，$????$\frac {3} {2})$??
??$(2)$??若??$∠PBA=90°$??
則??$P$??點(diǎn)的橫坐標(biāo)為??$4，$??代入??$y=\frac {1} {2}x+\frac {5} {2}$??
解得：??$y=\frac {9} {2}，$??則??$P(4，$????$\frac {9} {2})$??
??$(3)$??若??$∠APB=90°，$??則設(shè)??$P(x，$????$\frac 12x+\frac {5}{2})$??
在直角三角形中??$PA^2+PB^2=AB^2$??
∴??$(x+2)^2+(\frac x{2}+\frac 52)^2+(x-4)^2+(\frac x{2}+\frac 52)^2=6^2$??
??$x_1= 1 ，$????$x_2=-\frac {7}{5}$??
∴??$P(1，$????$3)$??或??$P(-\frac {7}{5}，$????$\frac {9}{5})$??
綜上所述點(diǎn)??$P$??的坐標(biāo)為：??$(-2，$????$\frac {3} {2})，$????$(4，$????$\frac 92)，$????$(1，$????$3)，$????$(-\frac {7}{5}，$????$\frac {9}{5})$??
解：①以?$OA$?為公共邊
?$B_1(0，$??$-3)、$??$B_2(4，$??$-3)、$??$B_3(4，$??$3)$?
②以?$OB$?為公共邊
?$A_1(-4，$??$0)、$??$A_2(-4，$??$3)、$??$A_3(4，$??$3)$?
③以?$AB$?為公共邊
設(shè)未知頂點(diǎn)坐標(biāo)為?$C(m，$??$n)$?
∵?$∠BCA=90°$?
∴點(diǎn)?$C$?到線段?$ABDE$?中點(diǎn)?$(2，$??$\frac 32)$?的距離為?$\frac 12AB=\frac 52$?
∴?$(m-2)^2+(n-\frac 32)^2=(\frac 52)^2$?
化簡得?$\mathrm {m^2}+n^2=4m+3n①$?
當(dāng)?$BC=4，$??$AC=3$?時
?$\begin{cases}{(m-0)^2+(n-3)^2=4^2}\\{(m-4)^2+(n-0)^2=3^2}\end{cases}$?
化簡之后得?$4m-3n=7，$?∴?$n=\frac {4m-7}3$?
將?$n=\frac {4m-7}3$?代入①解方程得?$m_1=4，$??$m_2=\frac {28}{25}$?
∴?$C(4，$??$3)$?或?$C(\frac {28}{25}，$??$-\frac {21}{25})$?
當(dāng)?$BC=3，$??$AC=4$?時
?$\begin{cases}{(m-0)^2+(n-3)^2=3^2}\\{(m-4)^2+(n-0)^2=4^2}\end{cases}$?
化簡之后得?$6n=8m，$??$n=\frac 43m$?
將?$n=\frac 43m$?代入①解方程得?$m_3=0，$??$m_4=\frac {72}{25}$?
當(dāng)?$m_3=0$?時，點(diǎn)?$C$?與點(diǎn)?$O$?重合，故舍去
∴?$C(\frac {72}{25}，$??$\frac {96}{25})$?
綜上所述，這個直角三角形的未知頂點(diǎn)坐標(biāo)為?$(0，$??$-3)、$??$(4，$??$-3)、$??$(4，$??$3)、$?
?$(-4，$??$0)、$??$(-4，$??$3)、$??$(\frac {28}{25}，$??$-\frac {21}{25})、$??$(\frac {72}{25}，$??$\frac {96}{25})$?
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