《2025年大联考单元期末测试卷八年级数学上册沪科版》

22. 如图，已知直线$m:y = \frac{3}{4}x + \frac{9}{4}$与$x$轴交于点$A$，与$y$轴交于点$B$，直线$n:y = -\frac{1}{2}x + b$与$x$轴交于点$C$，与$y$轴交于点$D$，直线$m$与直线$n$交于点$G(a,3)$。
（1）求直线$n$的函数表达式。
（2）连接$OG$，求$\triangle ODG$的面积。
（3）若点$E$在直线$m$上，且使得$S_{\triangle EDC} = S_{\triangle ODC}$，求点$E$的坐标。

答案： 22.解:
(1)$\because$直线$m$与直线$n$交于点$G(a,3)$,
　$\therefore 3 = \frac{3}{4}a + \frac{9}{4}$,解得$a = 1$,
　$\therefore$点$G$的坐标为$(1,3)$.
　把点$G(1,3)$代入直线$n:y = -\frac{1}{2}x + b$中,
　得$3 = -\frac{1}{2}+b$,解得$b = \frac{7}{2}$,
　$\therefore$直线$n$的函数表达式为$y = -\frac{1}{2}x + \frac{7}{2}$.(4分)
(2)令$x = 0$,则$y = -\frac{1}{2}x + \frac{7}{2}=\frac{7}{2}$,
　$\therefore$点$D$的坐标为$(0,\frac{7}{2}),\therefore OD = \frac{7}{2}$,
　$\therefore S_{\triangle ODG} = \frac{1}{2}×\frac{7}{2}×1 = \frac{7}{4}$.(7分)
(3)$\because S_{\triangle EDC} = S_{\triangle ODG},\therefore$点$E$到直线$n$的距离等于点$O$到直线$n$的距离.
　当点$E$在直线$n$的下方时,直线$OE$与直线$n$平行,
　$\therefore$直线$OE$的函数表达式为$y = -\frac{1}{2}x$.
　联立,得$\begin{cases}y = -\frac{1}{2}x\\y = \frac{3}{4}x + \frac{9}{4}\end{cases}$,解得$\begin{cases}x = -\frac{9}{5}\\y = \frac{9}{10}\end{cases}$,
　$\therefore$点$E$的坐标为$(-\frac{9}{5},\frac{9}{10})$.(9分)
　当点$E$在直线$n$的上方时,则过点$E$与直线$n$平行的直线过点$(0,7)$,
　$\therefore$该直线的函数表达式为$y = -\frac{1}{2}x + 7$.
　联立,得$\begin{cases}y = -\frac{1}{2}x + 7\\y = \frac{3}{4}x + \frac{9}{4}\end{cases}$,解得$\begin{cases}x = \frac{19}{5}\\y = \frac{51}{10}\end{cases}$,
　$\therefore$此时点$E$的坐标为$(\frac{19}{5},\frac{51}{10})$.
　综上所述,点$E$的坐标为$(-\frac{9}{5},\frac{9}{10})$或$(\frac{19}{5},\frac{51}{10})$(12分)