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1. 如图，已知直线$l_1$经过点$A(5,0)$，$B(1,4)$，与直线$l_2:y = 2x - 4$交于点$C$，直线$l_2$交$x$轴于点$D$.
(1)求直线$l_1$的解析式；
(2)求直线$l_1$与$l_2$的交点$C$的坐标；
(3)求$\triangle ACD$的面积.
(1)求直线$l_1$的解析式；
(2)求直线$l_1$与$l_2$的交点$C$的坐标；
(3)求$\triangle ACD$的面积.
答案：
(1)解:设直线$l_{1}$的解析式为$y = kx + b$.将$A(5,0)$,$B(1,4)$代入$y = kx + b$,得$\begin{cases}5k + b = 0,\\k + b = 4,\end{cases}$解得$\begin{cases}k = - 1,\\b = 5.\end{cases}$所以直线$l_{1}$的解析式为$y = - x + 5$.
(2)联立$\begin{cases}y = - x + 5,\\y = 2x - 4,\end{cases}$解得$\begin{cases}x = 3,\\y = 2.\end{cases}$所以点$C$的坐标为$(3,2)$.
(3)在$y = 2x - 4$中,令$y = 0$,则$2x - 4 = 0$,解得$x = 2$,所以点$D$的坐标为$(2,0)$.因为$A(5,0)$,所以$AD = 5 - 2 = 3$,所以$S_{\triangle ACD}=\frac{1}{2}AD\cdot|y_{C}|=\frac{1}{2}\times3\times2 = 3$.
(1)解:设直线$l_{1}$的解析式为$y = kx + b$.将$A(5,0)$,$B(1,4)$代入$y = kx + b$,得$\begin{cases}5k + b = 0,\\k + b = 4,\end{cases}$解得$\begin{cases}k = - 1,\\b = 5.\end{cases}$所以直线$l_{1}$的解析式为$y = - x + 5$.
(2)联立$\begin{cases}y = - x + 5,\\y = 2x - 4,\end{cases}$解得$\begin{cases}x = 3,\\y = 2.\end{cases}$所以点$C$的坐标为$(3,2)$.
(3)在$y = 2x - 4$中,令$y = 0$,则$2x - 4 = 0$,解得$x = 2$,所以点$D$的坐标为$(2,0)$.因为$A(5,0)$,所以$AD = 5 - 2 = 3$,所以$S_{\triangle ACD}=\frac{1}{2}AD\cdot|y_{C}|=\frac{1}{2}\times3\times2 = 3$.
2. 如图，已知直线$y = -\frac{1}{2}x + 2$交$x$轴于点$A$，交$y$轴于点$B$，点$C$的坐标为$(-1,0)$，连接$BC$.
(1)求$\triangle ABC$的面积；
(2)在直线$AB$上是否存在点$P$，使得$S_{\triangle POA}=2S_{\triangle BOA}$？若存在，请求出点$P$的坐标；若不存在，请说明理由.
(1)求$\triangle ABC$的面积；
(2)在直线$AB$上是否存在点$P$，使得$S_{\triangle POA}=2S_{\triangle BOA}$？若存在，请求出点$P$的坐标；若不存在，请说明理由.
答案：
(1)解:当$x = 0$时,$y = 2$,所以$B(0,2)$;当$y = 0$时,$-\frac{1}{2}x + 2 = 0$,解得$x = 4$,所以$A(4,0)$.因为$C(-1,0)$,所以$AC = 5$,$OB = 2$,所以$S_{\triangle ABC}=\frac{1}{2}AC\cdot OB=\frac{1}{2}\times5\times2 = 5$.
(2)存在.由
(1)易知$OA = 4$,$OB = 2$,所以$S_{\triangle BOA}=\frac{1}{2}\times4\times2 = 4$.因为$S_{\triangle POA}=2S_{\triangle BOA}$,所以$S_{\triangle POA}=8$,所以$\frac{1}{2}OA\cdot|y_{p}| = 8$,即$\frac{1}{2}\times4\times|y_{p}| = 8$,所以$y_{p}=\pm4$.把$y = 4$代入$y = -\frac{1}{2}x + 2$,得$-\frac{1}{2}x + 2 = 4$,解得$x = - 4$;把$y = - 4$代入$y = -\frac{1}{2}x + 2$,得$-\frac{1}{2}x + 2 = - 4$,解得$x = 12$.所以点$P$的坐标为$(-4,4)$或$(12,-4)$.
(1)解:当$x = 0$时,$y = 2$,所以$B(0,2)$;当$y = 0$时,$-\frac{1}{2}x + 2 = 0$,解得$x = 4$,所以$A(4,0)$.因为$C(-1,0)$,所以$AC = 5$,$OB = 2$,所以$S_{\triangle ABC}=\frac{1}{2}AC\cdot OB=\frac{1}{2}\times5\times2 = 5$.
(2)存在.由
(1)易知$OA = 4$,$OB = 2$,所以$S_{\triangle BOA}=\frac{1}{2}\times4\times2 = 4$.因为$S_{\triangle POA}=2S_{\triangle BOA}$,所以$S_{\triangle POA}=8$,所以$\frac{1}{2}OA\cdot|y_{p}| = 8$,即$\frac{1}{2}\times4\times|y_{p}| = 8$,所以$y_{p}=\pm4$.把$y = 4$代入$y = -\frac{1}{2}x + 2$,得$-\frac{1}{2}x + 2 = 4$,解得$x = - 4$;把$y = - 4$代入$y = -\frac{1}{2}x + 2$,得$-\frac{1}{2}x + 2 = - 4$,解得$x = 12$.所以点$P$的坐标为$(-4,4)$或$(12,-4)$.
3. 如图，直线$y = 2x + 4$与$x$轴交于点$A$，与$y$轴交于点$B$，$C$是$OB$的中点.
(1)求点$C$的坐标；
(2)若在$x$轴上存在一点$P$，使得$\triangle ABP$是直角三角形，请求出点$P$的坐标.
(1)求点$C$的坐标；
(2)若在$x$轴上存在一点$P$，使得$\triangle ABP$是直角三角形，请求出点$P$的坐标.
答案：
(1)解:在$y = 2x + 4$中,当$x = 0$时,$y = 4$.所以$B(0,4)$.因为$C$是$OB$的中点,所以易得点$C$的坐标为$(0,2)$.
(2)在$y = 2x + 4$中,当$y = 0$时,$x = - 2$,所以$A(-2,0)$.易得$OA = 2$,$OB = 4$.设点$P$的坐标为$(m,0)$,分两种情况讨论:①当$\angle APB = 90^{\circ}$时,点$P$与原点重合,此时点$P$的坐标为$(0,0)$; ②当$\angle ABP = 90^{\circ}$时,由勾股定理得$AP^{2}=AB^{2}+BP^{2}$.因为$AP = |m + 2|$,$AB^{2}=OA^{2}+OB^{2}=2^{2}+4^{2}=20$,$BP^{2}=OB^{2}+OP^{2}=4^{2}+m^{2}$,所以$(m + 2)^{2}=20 + 4^{2}+m^{2}$,解得$m = 8$.所以$P(8,0)$.综上所述,点$P$的坐标为$(0,0)$或$(8,0)$.
(1)解:在$y = 2x + 4$中,当$x = 0$时,$y = 4$.所以$B(0,4)$.因为$C$是$OB$的中点,所以易得点$C$的坐标为$(0,2)$.
(2)在$y = 2x + 4$中,当$y = 0$时,$x = - 2$,所以$A(-2,0)$.易得$OA = 2$,$OB = 4$.设点$P$的坐标为$(m,0)$,分两种情况讨论:①当$\angle APB = 90^{\circ}$时,点$P$与原点重合,此时点$P$的坐标为$(0,0)$; ②当$\angle ABP = 90^{\circ}$时,由勾股定理得$AP^{2}=AB^{2}+BP^{2}$.因为$AP = |m + 2|$,$AB^{2}=OA^{2}+OB^{2}=2^{2}+4^{2}=20$,$BP^{2}=OB^{2}+OP^{2}=4^{2}+m^{2}$,所以$(m + 2)^{2}=20 + 4^{2}+m^{2}$,解得$m = 8$.所以$P(8,0)$.综上所述,点$P$的坐标为$(0,0)$或$(8,0)$.
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