10. 如图,一个点在第一象限及$x$轴、$y$轴上移动,在第$1$秒,它从原点移动到点$(1,0)$,然后按照图中箭头所示方向移动,即$(0,0)\to(1,0)\to(1,1)\to(0,1)\to(0,2)\to\cdots$,且每秒移动一个单位长度,那么第$2020$秒时,点所在位置的坐标是 (
A. $(64,44)$
B. $(45,5)$
C. $(44,5)$
D. $(44,4)$
D
)A. $(64,44)$
B. $(45,5)$
C. $(44,5)$
D. $(44,4)$
答案：D
11. (2023春·海安期末)在平面直角坐标系中,点$A(-m,0)$,$B(4-m,0)$,$C(m,2)$,$D(m+6,2)$,若线段$CD$上存在点$E$,过点$E$作$EF\perp AB$,垂足为$F$,点$F$恰好是线段$AB$的中点,则实数$m$的取值范围是
-2≤m≤1
.答案：- 2≤m≤1
12. 如图,在平面直角坐标系中,$\triangle ABC$为直角三角形,$A$,$C$两点分别在$x$轴、$y$轴上,$\angle B=90^{\circ}$,$B$点的坐标为$(1,3)$.将$\triangle ABC$沿$AC$翻折,$B$点落在$D$点位置,$AD$交$y$轴于点$E$,求点$D$的坐标.
(
(
(-\frac{4}{5},\frac{12}{5})
)答案：解:如答图，过点D作DH⊥OC于点H.
∵点B的坐标为(1,3),∴AO = 1,AB = 3.
根据折叠可知CD = CB = OA,
而∠CDA = ∠AOE = 90°,∠DEC = ∠AEO,∴△CDE≌△AOE(AAS),
∴OE = DE,OA = CD = 1.
设OE = x,那么CE = 3 - x,DE = x,
∴在Rt△DCE中,CE² = DE² + CD²,
∴(3 - x)² = x² + 1²,∴x = $\frac{4}{3}$,∴CE = $\frac{5}{3}$,DE = $\frac{4}{3}$.
又∵DH⊥CE,∴$\frac{1}{2}$CE×DH = $\frac{1}{2}$CD×DE,
∴DH = $\frac{CD\cdot DE}{CE}$ = $\frac{4}{5}$,
在Rt△CDH中，$CH=\sqrt{CD^{2}-DH^{2}}=\sqrt{1^{2}-(\frac{4}{5})^{2}}=\frac{3}{5}$,
∴OH = 3 - $\frac{3}{5}$ = $\frac{12}{5}$.
∵点D在第二象限，∴点D的坐标为( - $\frac{4}{5}$,$\frac{12}{5}$).
∵点B的坐标为(1,3),∴AO = 1,AB = 3.
根据折叠可知CD = CB = OA,
而∠CDA = ∠AOE = 90°,∠DEC = ∠AEO,∴△CDE≌△AOE(AAS),
∴OE = DE,OA = CD = 1.
设OE = x,那么CE = 3 - x,DE = x,
∴在Rt△DCE中,CE² = DE² + CD²,
∴(3 - x)² = x² + 1²,∴x = $\frac{4}{3}$,∴CE = $\frac{5}{3}$,DE = $\frac{4}{3}$.
又∵DH⊥CE,∴$\frac{1}{2}$CE×DH = $\frac{1}{2}$CD×DE,
∴DH = $\frac{CD\cdot DE}{CE}$ = $\frac{4}{5}$,
在Rt△CDH中，$CH=\sqrt{CD^{2}-DH^{2}}=\sqrt{1^{2}-(\frac{4}{5})^{2}}=\frac{3}{5}$,
∴OH = 3 - $\frac{3}{5}$ = $\frac{12}{5}$.
∵点D在第二象限，∴点D的坐标为( - $\frac{4}{5}$,$\frac{12}{5}$).
13. 如图,在平面直角坐标系中,$A(0,a)$,$B(b,0)$,且$a$,$b$满足$\begin{cases}2a - b = 20,\\3a + 2b = 2.\end{cases}$作射线$BA$,动点$P$从点$B$开始沿射线$BA$以每秒$2$个单位长度的速度运动,设运动时间为$t$秒.
(1)求点$A$,$B$的坐标;
(2)设$\triangle AOP$的面积为$S$,用含$t$的式子表示$S$,并直接写出$t$的取值范围;
(3)$M$为线段$OP$的中点,连接$AM$,当点$P$在线段$BA$上,$\triangle AOM$的面积为$\triangle AOB$的面积的$\frac{1}{3}$时,求出$t$的值,并求出此时点$M$到$x$轴的距离.
(1)求点$A$,$B$的坐标;
(2)设$\triangle AOP$的面积为$S$,用含$t$的式子表示$S$,并直接写出$t$的取值范围;
(3)$M$为线段$OP$的中点,连接$AM$,当点$P$在线段$BA$上,$\triangle AOM$的面积为$\triangle AOB$的面积的$\frac{1}{3}$时,求出$t$的值,并求出此时点$M$到$x$轴的距离.
答案：
解:(1)由$\begin{cases}2a - b = 20\\3a + 2b = 2\end{cases}$解得$\begin{cases}a = 6\\b = - 8\end{cases}$
∴A(0,6),B( - 8,0).
(2)如答图①,作OH⊥AB于点H.∵$S_{\triangle AOB}=\frac{1}{2}\cdot OA\cdot OB=\frac{1}{2}\cdot AB\cdot OH$,∴$OH=\frac{6×8}{10}=\frac{24}{5}$.
∵AB = 10,∴点P从点B运动到点A的时间为5秒，当0≤t＜5时,$S=\frac{1}{2}\cdot PA\cdot OH=\frac{1}{2}(10 - 2t)×\frac{24}{5}=24-\frac{24}{5}t$.
当t＞5时,$S=\frac{1}{2}\cdot PA\cdot OH=\frac{1}{2}(2t - 10)×\frac{24}{5}=\frac{24}{5}t - 24$.
综上所述,$S=\begin{cases}24-\frac{24}{5}t(0\leqslant t＜5)\\\frac{24}{5}t - 24(t＞5)\end{cases}$
(3)如答图②,连接BM.
∵OM = PM,∴$S_{\triangle AOM}=S_{\triangle APM}$,
∵$S_{\triangle AOM}=\frac{1}{3}S_{\triangle AOB}$,∴$S_{\triangle OPB}=\frac{1}{3}S_{\triangle AOB}$,
∴BP = $\frac{1}{3}$AB,∴2t = $\frac{10}{3}$,即t = $\frac{5}{3}$.
设点M到x轴的距离为h.
∵OM = PM,∴$S_{\triangle OBM}=\frac{1}{2}S_{\triangle OPB}=\frac{1}{6}S_{\triangle AOB}$,
∴$\frac{1}{2}$×8×h = $\frac{1}{6}$×$\frac{1}{2}$×6×8,
解得h = 1,∴点M到x轴的距离为1.
解:(1)由$\begin{cases}2a - b = 20\\3a + 2b = 2\end{cases}$解得$\begin{cases}a = 6\\b = - 8\end{cases}$
∴A(0,6),B( - 8,0).
(2)如答图①,作OH⊥AB于点H.∵$S_{\triangle AOB}=\frac{1}{2}\cdot OA\cdot OB=\frac{1}{2}\cdot AB\cdot OH$,∴$OH=\frac{6×8}{10}=\frac{24}{5}$.
∵AB = 10,∴点P从点B运动到点A的时间为5秒，当0≤t＜5时,$S=\frac{1}{2}\cdot PA\cdot OH=\frac{1}{2}(10 - 2t)×\frac{24}{5}=24-\frac{24}{5}t$.
当t＞5时,$S=\frac{1}{2}\cdot PA\cdot OH=\frac{1}{2}(2t - 10)×\frac{24}{5}=\frac{24}{5}t - 24$.
综上所述,$S=\begin{cases}24-\frac{24}{5}t(0\leqslant t＜5)\\\frac{24}{5}t - 24(t＞5)\end{cases}$
(3)如答图②,连接BM.
∵OM = PM,∴$S_{\triangle AOM}=S_{\triangle APM}$,
∵$S_{\triangle AOM}=\frac{1}{3}S_{\triangle AOB}$,∴$S_{\triangle OPB}=\frac{1}{3}S_{\triangle AOB}$,
∴BP = $\frac{1}{3}$AB,∴2t = $\frac{10}{3}$,即t = $\frac{5}{3}$.
设点M到x轴的距离为h.
∵OM = PM,∴$S_{\triangle OBM}=\frac{1}{2}S_{\triangle OPB}=\frac{1}{6}S_{\triangle AOB}$,
∴$\frac{1}{2}$×8×h = $\frac{1}{6}$×$\frac{1}{2}$×6×8,
解得h = 1,∴点M到x轴的距离为1.