《2025年名校课堂八年级数学上册沪科版》

2. (2024·合肥168中期末)如图，已知直线$l_1:y = kx - 2$与直线$y = x$平行，与$x$轴交于点$A$，与$y$轴交于点$B$.直线$l_2$与$y$轴交于点$C(0,4)$，与$x$轴交于点$D$，与直线$l_1$交于点$E(3,m)$.
(1)求直线$l_2$的表达式；
(2)求四边形$AOCE$的面积.

答案：
(1)$\because$直线$l_1:y = kx - 2$与直线$y = x$平行，$\therefore k = 1$．$\therefore$直线$l_1$的表达式为$y = x - 2$．$\because$点$E(3,m)$在直线$l_1$上，$\therefore m = 3 - 2 = 1$．$\therefore E(3,1)$．设直线$l_2$的表达式为$y = ax + b$．把$C(0,4)$，$E(3,1)$代入，得$\begin{cases} b = 4, \\3a + b = 1,\end{cases}$解得$\begin{cases} a = - 1, \\b = 4.\end{cases}$$\therefore$直线$l_2$的表达式为$y = - x + 4$．
(2)在直线$l_1:y = x - 2$中，令y=0，则$x - 2 = 0$，解得$x = 2$．$\therefore A(2,0)$．在直线$l_2:y = - x + 4$中，令y=0，则$- x + 4 = 0$，解得$x = 4$．$\therefore D(4,0)$．$\therefore S_{\triangle COD} = \frac{1}{2} × 4 × 4 = 8$，$S_{\triangle AED} = \frac{1}{2} × (4 - 2) × 1 = 1$．$\therefore S_{四边形AOCE} = S_{\triangle COD} - S_{\triangle AED} = 8 - 1 = 7$．

3. (2024·六安裕安区期末)如图，一次函数$y = -\frac{4}{3}x + 4$的图象分别与$x$轴，$y$轴的正半轴交于点$A$，$B$，一次函数$y = kx - 4$的图象与直线$AB$交于点$C(m,2)$，且交$x$轴于点$D$.
(1)求$m$的值及点$A$，$B$的坐标；
(2)求三角形$ACD$的面积；
(3)若$P$是$x$轴上的一个动点，当$S_{三角形PCD} = \frac{1}{2}S_{三角形ACD}$时，求点$P$的坐标.

答案：
(1)$\because$一次函数$y = - \frac{4}{3}x + 4$的图象经过点$C(m,2)$，$\therefore - \frac{4}{3}m + 4 = 2$，解得$m = \frac{3}{2}$．$\because$一次函数$y = - \frac{4}{3}x + 4$的图象分别与x轴，y轴的正半轴交于点A，B，$\therefore$当y=0时，$- \frac{4}{3}x + 4 = 0$，解得$x = 3$．$\therefore A(3,0)$．当x=0时，y=4，$\therefore B(0,4)$．
(2)把点$C(\frac{3}{2},2)$代入$y = kx - 4$，得$2 = \frac{3}{2}k - 4$，解得$k = 4$．$\therefore$直线CD的表达式为$y = 4x - 4$．当y=0时，x=1，即$D(1,0)$．$\therefore AD = 3 - 1 = 2$．$\therefore S_{\triangle ACD} = \frac{1}{2} × 2 × 2 = 2$．
(3)设$P(a,0)$．$\therefore PD = |a - 1|$．$\because S_{\triangle PCD} = \frac{1}{2}S_{\triangle ACD}$，$\therefore \frac{1}{2} × |a - 1| × 2 = \frac{1}{2} × 2$．$\therefore a = 2$或$a = 0$．$\therefore$点P的坐标为(2,0)或(0,0)．