《2025年通城学典活页检测九年级数学下册人教版》

5. 如图，一次函数$y_{1}=kx + b$的图象与反比例函数$y_{2}=\frac{m}{x}$的图象交于点$A(-2,-5)$，$C(5,n)$，交$y$轴于点$B$，交$x$轴于点$D$. 连接$OA$，$OC$，则$\triangle AOC$的面积为______.

答案： $\frac{21}{2}$ 解析：把$A(-2,-5)$代入$y_{2}=\frac{m}{x}$，得$m = - 2\times(-5)=10$，$\therefore$反比例函数的解析式为$y_{2}=\frac{10}{x}$.把$C(5,n)$代入$y_{2}=\frac{10}{x}$，得$n=\frac{10}{5}=2$.$\therefore C(5,2)$.把$A(-2,-5)$，$C(5,2)$代入$y_{1}=kx + b$，得$\begin{cases}-2k + b = -5\\5k + b = 2\end{cases}$，解得$\begin{cases}k = 1\\b = -3\end{cases}$.$\therefore$一次函数的解析式为$y_{1}=x - 3$.在$y_{1}=x - 3$中，当$y_{1}=0$时，$x - 3 = 0$，解得$x = 3$，$\therefore D(3,0)$.$\therefore OD = 3$.$\therefore S_{\triangle AOC}=S_{\triangle COD}+S_{\triangle AOD}=\frac{1}{2}\times3\times2+\frac{1}{2}\times3\times5=\frac{21}{2}$.

6. ★（雅安中考）如图，在直角坐标系中，一次函数的图象$l$与反比例函数$y=\frac{k}{x}$的图象交于$M(\frac{1}{2},4)$，$N(n,1)$两点.
（1）求反比例函数及一次函数的解析式.
（2）求$\triangle OMN$的面积.
（3）若$P$是$y$轴上一动点，连接$PM$，$PN$. 当$PM + PN$的值最小时，求点$P$的坐标.

答案：

(1)$\because$点$M(\frac{1}{2},4)$在反比例函数$y=\frac{k}{x}$的图象上，$\therefore k=\frac{1}{2}\times4 = 2$.$\therefore$反比例函数的解析式为$y=\frac{2}{x}$.$\because$点$N(n,1)$在反比例函数$y=\frac{2}{x}$的图象上，$\therefore n = 2$.$\therefore N(2,1)$.设一次函数的解析式为$y = ax + b$.把$M(\frac{1}{2},4)$，$N(2,1)$代入，得$\begin{cases}\frac{1}{2}a + b = 4\\2a + b = 1\end{cases}$，解得$\begin{cases}a = - 2\\b = 5\end{cases}$.$\therefore$一次函数的解析式为$y = - 2x + 5$.
(2)如图，设直线$l$交$x$轴于点$A$，交$y$轴于点$B$.在$y = - 2x + 5$中，令$x = 0$，则$y = 5$；令$y = 0$，则$x=\frac{5}{2}$，$\therefore A(\frac{5}{2},0)$，$B(0,5)$.$\therefore OA=\frac{5}{2}$，$OB = 5$.$\therefore S_{\triangle OMN}=S_{\triangle AOB}-S_{\triangle AON}-S_{\triangle BOM}=\frac{1}{2}OA\cdot OB-\frac{1}{2}OA\cdot y_{N}-\frac{1}{2}OB\cdot x_{M}=\frac{1}{2}\times\frac{5}{2}\times5-\frac{1}{2}\times\frac{5}{2}\times1-\frac{1}{2}\times5\times\frac{1}{2}=\frac{15}{4}$.
(3)如图，作点$M$关于$y$轴的对称点$M'$，连接$M'N$交$y$轴于点$P$，则$M'N$的长即为$PM + PN$的最小值.$\because$点$M(\frac{1}{2},4)$与点$M'$关于$y$轴对称，$\therefore M'(-\frac{1}{2},4)$.设直线$M'N$对应的函数解析式为$y = cx + d$.把$M'(-\frac{1}{2},4)$，$N(2,1)$代入，得$\begin{cases}-\frac{1}{2}c + d = 4\\2c + d = 1\end{cases}$，解得$\begin{cases}c = -\frac{6}{5}\\d=\frac{17}{5}\end{cases}$.$\therefore$直线$M'N$对应的函数解析式为$y = -\frac{6}{5}x+\frac{17}{5}$.令$x = 0$，则$y=\frac{17}{5}$.$\therefore$点$P$的坐标为$(0,\frac{17}{5})$.

7. ★★（泸州中考）如图，在直角坐标系中，一次函数$y = kx + b$的图象与$x$轴相交于点$A(-2,0)$，与反比例函数$y=\frac{a}{x}$的图象相交于点$B(2,3)$.
（1）求一次函数和反比例函数的解析式；
（2）直线$x = m(m＞2)$与反比例函数$y=\frac{a}{x}(x＞0)$和$y=-\frac{2}{x}(x＞0)$的图象分别交于点$C$，$D$，且$S_{\triangle OBC}=2S_{\triangle OCD}$，求点$C$的坐标.

答案：

(1)将$A(-2,0)$，$B(2,3)$代入$y = kx + b$，得$\begin{cases}-2k + b = 0\\2k + b = 3\end{cases}$，解得$\begin{cases}k=\frac{3}{4}\\b=\frac{3}{2}\end{cases}$.$\therefore$一次函数的解析式为$y=\frac{3}{4}x+\frac{3}{2}$.将$B(2,3)$代入$y=\frac{a}{x}$，得$a = 2\times3 = 6$，$\therefore$反比例函数的解析式为$y=\frac{6}{x}$.
(2)将$x = m$分别代入$y=\frac{6}{x}$和$y = -\frac{2}{x}$，得点$C$的坐标为$(m,\frac{6}{m})$，点$D$的坐标为$(m,-\frac{2}{m})$.$\therefore S_{\triangle OCD}=\frac{1}{2}[\frac{6}{m}-(-\frac{2}{m})]\cdot m = 4$.又$\because S_{\triangle OBC}=2S_{\triangle OCD}$，$\therefore S_{\triangle OBC}=8$.如图，令直线$CD$与$x$轴的交点为$M$，过点$B$作$x$轴的垂线，垂足为$N$.$\because S_{\triangle BMN}+S_{梯形BNMC}=S_{\triangle OBC}+S_{\triangle OCM}$，且$S_{\triangle BON}=S_{\triangle OCM}$，$\therefore S_{梯形BNMC}=S_{\triangle OBC}=8$.$\therefore\frac{(\frac{6}{m}+3)(m - 2)}{2}=8$，解得$m_{1}=6$，$m_{2}=-\frac{2}{3}$.$\because m＞2$，$\therefore m = 6$.$\therefore$点$C$的坐标为$(6,1)$.