22. (14分)在平面直角坐标系xOy中,点P的坐标为$(a,b)$,实数a,b,m满足以下两个等式:$2a-m+1= 0,3b-2m+5= 0$.
(1)当$a= 1$时,点P到x轴的距离为
(2)若点P落在x轴上,求点P的坐标;
(1)当$a= 1$时,点P到x轴的距离为
$\frac{1}{3}$
;(2)若点P落在x轴上,求点P的坐标;
解:$\because$点$P(a,b)$落在$x$轴上,$\therefore b = 0$,
$\therefore \begin{cases}2a - m + 1 = 0,\\0 - 2m + 5 = 0,\end{cases}$解得$\begin{cases}m = \frac{5}{2},\\a = \frac{3}{4},\end{cases}\therefore P(\frac{3}{4},0)$.
(3)当$a\leqslant4\lt b$时,求m的整数值.$\therefore \begin{cases}2a - m + 1 = 0,\\0 - 2m + 5 = 0,\end{cases}$解得$\begin{cases}m = \frac{5}{2},\\a = \frac{3}{4},\end{cases}\therefore P(\frac{3}{4},0)$.
解:$\because 2a - m + 1 = 0,3b - 2m + 5 = 0$,
$\therefore a = \frac{m - 1}{2},b = \frac{2m - 5}{3}$,
$\because a \leq 4 ＜ b,\therefore \begin{cases}\frac{m - 1}{2} \leq 4,\frac{2m - 5}{3} ＞ 4,\end{cases}$解得$\frac{17}{2} ＜ m \leq 9$,
$\because m$是整数,$\therefore m = 9$.
$\therefore a = \frac{m - 1}{2},b = \frac{2m - 5}{3}$,
$\because a \leq 4 ＜ b,\therefore \begin{cases}\frac{m - 1}{2} \leq 4,\frac{2m - 5}{3} ＞ 4,\end{cases}$解得$\frac{17}{2} ＜ m \leq 9$,
$\because m$是整数,$\therefore m = 9$.
答案：(1)$\frac{1}{3}$
(2)解:$\because$点$P(a,b)$落在$x$轴上,$\therefore b = 0$,
$\therefore \begin{cases}2a - m + 1 = 0,\\0 - 2m + 5 = 0,\end{cases}$解得$\begin{cases}m = \frac{5}{2},\\a = \frac{3}{4},\end{cases}\therefore P(\frac{3}{4},0)$.
(3)解:$\because 2a - m + 1 = 0,3b - 2m + 5 = 0$,
$\therefore a = \frac{m - 1}{2},b = \frac{2m - 5}{3}$,
$\because a \leq 4 ＜ b,\therefore \begin{cases}\frac{m - 1}{2} \leq 4,\frac{2m - 5}{3} ＞ 4,\end{cases}$解得$\frac{17}{2} ＜ m \leq 9$,
$\because m$是整数,$\therefore m = 9$.
(2)解:$\because$点$P(a,b)$落在$x$轴上,$\therefore b = 0$,
$\therefore \begin{cases}2a - m + 1 = 0,\\0 - 2m + 5 = 0,\end{cases}$解得$\begin{cases}m = \frac{5}{2},\\a = \frac{3}{4},\end{cases}\therefore P(\frac{3}{4},0)$.
(3)解:$\because 2a - m + 1 = 0,3b - 2m + 5 = 0$,
$\therefore a = \frac{m - 1}{2},b = \frac{2m - 5}{3}$,
$\because a \leq 4 ＜ b,\therefore \begin{cases}\frac{m - 1}{2} \leq 4,\frac{2m - 5}{3} ＞ 4,\end{cases}$解得$\frac{17}{2} ＜ m \leq 9$,
$\because m$是整数,$\therefore m = 9$.
23. (14分)如图,在平面直角坐标系中,$A(a,0),B(b,0),C(-1,2)$,且$|a+2|+(b-3)^2= 0$.
(1)求a,b的值.
(2)①如图①,在y轴的正半轴上存在一点M,使$S_{\triangle COM}= \frac{1}{2}S_{\triangle ABC}$,求点M的坐标;
②在坐标轴的其他位置是否存在点M,使$S_{\triangle COM}= \frac{1}{2}S_{\triangle ABC}$仍然成立? 若存在,请求出符合条件的点M的坐标.
(3)如图②,过点C作$CD\perp y$轴交y轴于点D,点P为线段CD延长线上一动点,连接OP,OE平分$\angle AOP,OF\perp OE$.当点P运动时,$\frac{\angle OPD}{\angle DOE}$的值是否会改变? 若不变,求其值;若改变,说明理由.
(1)求a,b的值.
(2)①如图①,在y轴的正半轴上存在一点M,使$S_{\triangle COM}= \frac{1}{2}S_{\triangle ABC}$,求点M的坐标;
②在坐标轴的其他位置是否存在点M,使$S_{\triangle COM}= \frac{1}{2}S_{\triangle ABC}$仍然成立? 若存在,请求出符合条件的点M的坐标.
(3)如图②,过点C作$CD\perp y$轴交y轴于点D,点P为线段CD延长线上一动点,连接OP,OE平分$\angle AOP,OF\perp OE$.当点P运动时,$\frac{\angle OPD}{\angle DOE}$的值是否会改变? 若不变,求其值;若改变,说明理由.
答案：
解: (1)$\because |a + 2| + (b - 3)^2 = 0,\therefore a = -2,b = 3$.
(2)①设$M(0,m)(m ＞ 0)$,由题意得$\frac{1}{2}m \cdot 1 = \frac{1}{2} × \frac{1}{2} × (2 + 3) × 2$,解得$m = 5,\therefore M(0,5)$.
②当点$M$在$y$轴的负半轴上时,$\frac{1}{2}(-m) \cdot 1 = \frac{1}{2} × \frac{1}{2} × (2 + 3) × 2$,解得$m = -5,\therefore M(0,-5)$;
当点$M$在$x$轴上时,设$M(n,0)$,则$\frac{1}{2} × |n| × 2 = \frac{1}{2} × \frac{1}{2} × (2 + 3) × 2$,解得$n = \pm 2.5,\therefore M(\pm 2.5,0)$.
$\therefore M(2.5,0)$或$M(-2.5,0)$或$M(0,-5)$.
(3)不变.
如答图,使点$E,F$在直线$CD$上.$\because OE \perp OF,CD \perp OD$,
$\therefore \angle OED + \angle EFO = 90^{\circ},\angle DOE + \angle DEO = 90^{\circ}$,
$\angle AOE + \angle FOB = 90^{\circ},\angle EOP + \angle POF = 90^{\circ},\therefore \angle EOD = \angle EFO,\because OE$平分$\angle AOP,EF // AB$,
$\therefore \angle AOE = \angle EOP,\angle OFE = \angle FOB,\therefore \angle FOP = \angle FOB = \angle OFP,\because \angle OPD = \angle PFO + \angle POF = 2\angle OFP = 2\angle DOE,\therefore \frac{\angle OPD}{\angle DOE} = 2$.
解: (1)$\because |a + 2| + (b - 3)^2 = 0,\therefore a = -2,b = 3$.
(2)①设$M(0,m)(m ＞ 0)$,由题意得$\frac{1}{2}m \cdot 1 = \frac{1}{2} × \frac{1}{2} × (2 + 3) × 2$,解得$m = 5,\therefore M(0,5)$.
②当点$M$在$y$轴的负半轴上时,$\frac{1}{2}(-m) \cdot 1 = \frac{1}{2} × \frac{1}{2} × (2 + 3) × 2$,解得$m = -5,\therefore M(0,-5)$;
当点$M$在$x$轴上时,设$M(n,0)$,则$\frac{1}{2} × |n| × 2 = \frac{1}{2} × \frac{1}{2} × (2 + 3) × 2$,解得$n = \pm 2.5,\therefore M(\pm 2.5,0)$.
$\therefore M(2.5,0)$或$M(-2.5,0)$或$M(0,-5)$.
(3)不变.
如答图,使点$E,F$在直线$CD$上.$\because OE \perp OF,CD \perp OD$,
$\therefore \angle OED + \angle EFO = 90^{\circ},\angle DOE + \angle DEO = 90^{\circ}$,
$\angle AOE + \angle FOB = 90^{\circ},\angle EOP + \angle POF = 90^{\circ},\therefore \angle EOD = \angle EFO,\because OE$平分$\angle AOP,EF // AB$,
$\therefore \angle AOE = \angle EOP,\angle OFE = \angle FOB,\therefore \angle FOP = \angle FOB = \angle OFP,\because \angle OPD = \angle PFO + \angle POF = 2\angle OFP = 2\angle DOE,\therefore \frac{\angle OPD}{\angle DOE} = 2$.