《2026年王朝霞考点梳理时习卷九年级数学全一册人教版》

20. [朝霞原创] (9分)如图,在平面直角坐标系中,菱形$ABCD$的边$BC$在$x$轴上,$AC$与$BD$交于点$E$,顶点$B$
的坐标为$(2,0),\angle ABO=60°$,反比例函数$y=\frac{k}{x} (x＞0)$的图象经过顶点$D$.
(1)求反比例函数的解析式;
(2)将菱形$ABCD$沿$x$轴向右平移,使点$E$落在反比例函数的图象上,求菱形$ABCD$向右右移的距离.

答案： 解:
(1)$\because$点$B ( 2 , 0 )$，$\therefore O B = 2$．
$\because \angle A B O = 6 0 ^ { \circ }$，$\angle A O B = 9 0 ^ { \circ }$，
$\therefore \angle O A B = 3 0 ^ { \circ }$．
$\therefore A B = 2 O B = 4$．$\therefore O A = \sqrt { A B ^ { 2 } - O B ^ { 2 } } = 2 \sqrt { 3 }$．
$\because$四边形$A B C D$为菱形，
$\therefore A B = A D = 4$．
$\therefore$点$D ( 4 , 2 \sqrt { 3 } )$．
把点$D ( 4 , 2 \sqrt { 3 } )$代入$y = \frac { k } { x }$，得$k = 8 \sqrt { 3 }$．
$\therefore$反比例函数的解析式为$y = \frac { 8 \sqrt { 3 } } { x } ( x ＞ 0 )$．
(2)$\because$四边形$A B C D$为菱形，$\therefore E$为$B D$的中点．
$\because$点$B ( 2 , 0 )$，点$D ( 4 , 2 \sqrt { 3 } )$，
$\therefore$点$E ( 3 , \sqrt { 3 } )$．
把$y = \sqrt { 3 }$代入$y = \frac { 8 \sqrt { 3 } } { x }$，得$x = 8$．
$\because 8 - 3 = 5$，$\therefore$当平移至点$E$落在反比例函数的图象上时，菱形$A B C D$向右平移的距离为$5$个单位长度．

21. [朝霞原创] (10分)如图,在平面直角坐标系中,$\triangle OAB$是直角三角形,点$A$在$x$轴上,$\angle OAB=90°$,点$B$的
坐标为$(6,8)$,反比例函数$y=\frac{k}{x} (x＞0)$的图象过$OB$的中点$C$,且与$AB$交于点$D$.
(1)求$k$的值及点$D$的坐标;
(2)请用无刻度的直尺和圆规作出$\angle AOB$的平分线,交反比例函数$y=\frac{k}{x} (x＞0)$的图象于点$M$;
(要求:不写作法,保留作图痕迹)
(3)求(2)中$OM$所在直线的函数解析式.

答案：
解:
(1)$\because$点$B$的坐标为$( 6 , 8 )$，点$C$为$O B$的中点，$\therefore$点$C$的坐标为$( 3 , 4 )$．
将点$C ( 3 , 4 )$代入$y = \frac { k } { x } ( x ＞ 0 )$，得$k = 3 × 4 = 1 2$．
$\therefore$反比例函数的解析式为$y = \frac { 1 2 } { x } ( x ＞ 0 )$．
$\because \angle O A B = 9 0 ^ { \circ }$，即$A B \perp x$轴，
$\therefore$点$D$的横坐标为$6$．
将$x = 6$代入$y = \frac { 1 2 } { x } ( x ＞ 0 )$，得$y = 2$．
$\therefore$点$D$的坐标为$( 6 , 2 )$．
(2)如图所示．

(3)过点$C$作$C E \perp y$轴于点$E$，反向延长$C E$交射线$O M$于点$F$，如图．
则$E F // O A$．$\therefore \angle C F O = \angle A O F$．
$\because$点$C ( 3 , 4 )$，$\therefore C E = 3$，$O E = 4$．
$\therefore$在$ Rt \triangle O E C$中，$O C = \sqrt { C E ^ { 2 } + O E ^ { 2 } } = 5$．
$\because O M$平分$\angle A O B$，$\therefore \angle A O F = \angle C O F$．
$\therefore \angle C F O = \angle C O F$．$\therefore C F = O C = 5$．
$\therefore E F = C F + C E = 8$．$\therefore$点$F$的坐标为$( 8 , 4 )$．
设$O M$所在直线的函数解析式为$y = m x$．
将点$F ( 8 , 4 )$代入$y = m x$，得$8 m = 4$．
解得$m = \frac { 1 } { 2 }$．
$\therefore O M$所在直线的函数解析式为$y = \frac { 1 } { 2 } x$．