18. (8分)在平面直角坐标系中,点$P(3a-14,$$2-a).$
(1)若点P在y轴上,求点P的坐标;
(2)若点P位于第三象限且横、纵坐标都是整数,求点P的坐标.

18. (1) 由题意得$3a-14=0$,解得$a=\dfrac{14}{3}$.$\therefore$ 点$P$的坐标为$(0,-\dfrac{8}{3})$.
(2)
∵ 点$P(3a-14,2-a)$位于第三象限,$\therefore 3a-14＜0$,$2-a＜0$,解得$2＜a＜\dfrac{14}{3}$.$\because$ 点$P$的横、纵坐标都是整数,$\therefore a=3$或$a=4$.当$a=3$时,点$P(-5,-1)$.当$a=4$时,点$P(-2,-2)$,$\therefore$ 点$P$的坐标为$(-5,-1)$或$(-2,-2)$.

19. (8分)在平面直角坐标系中,我们把横、纵坐标都为整数的点称为整点,记顶点都是整点的三角形为整点三角形. 如图,已知整点$A(2,2),B(3,5)$,请在所给的网格区域(含边界)按要求画整点三角形.
(1)在图①中画一个$△ ABC$,使点$C$的横、纵坐标的平方和等于25.
(2)在图②中画一个$△ OBD$,使点$D$的横、纵坐标之和等于4,且点$A$在$△ OBD$的内部.

19. 画法合理即可,如:(1) 如图①所示,$△ ABC$即为所求.
(2) 如图②所示,$△ OBD$即为所求.

20. (8分)(2026·枣庄期中)在平面直角坐标系$xOy$中,点$A$的坐标为$(2,4)$,过点$(3,0)$作$x$轴的垂线$l$,点$A$与点$B$关于直线$l$对称.
(1)点$B$的坐标为.
(2)点$C$的坐标为$(6,0)$,顺次连接$OABC$,若在四边形$OABC$内部有一个点$P$,满足$S_{△ POA}=S_{△ PBC}$,且$S_{△ PAB}=S_{△ POC}$,求点$P$的坐标.

20. (1) $(4,4)$ 解析:
∵ 点$A$的坐标为$(2,4)$,过点$(3,0)$作$x$轴的垂线$l$,$\therefore$ 点$A$到直线$x=3$的距离为1,则$B(4,4)$.

(2) 如图,$\because C(6,0)$,$O(0,0)$,$\therefore$ 点$C$与点$O$关于$l$对称.$\because$ 在四边形$OABC$内部有一个点$P$,满足$S_{△ POA}=S_{△ PBC}$,$\therefore$ 点$P$在$l$上,设点$P(3,p)$.$\because A(2,4)$,$B(4,4)$,$\therefore AB // OC$.$\because AB=2$,$OC=6$,且$S_{△ PAB}=S_{△ POC}$,$\therefore \dfrac{1}{2}AB × |y_A-y_P|=\dfrac{1}{2}OC × |y_P|$,即$|4-p|=3|p|$,解得$p=1$或$p=-2$.$\because P$在四边形$OABC$内部,$\therefore p=1$.$\therefore$ 点$P$的坐标为$(3,1)$.

21. (10 分)(2026·成都期中) 如图,已知点 A的坐标是$(4,2)$,其中点 C,B 分别在 x 轴和y 轴上,四边形 ABOC 为长方形,连接 OA,将$△ AOC$沿对角线 OA 折叠,使点 C 落在$C'$处,$OC'$交 AB 于点 D.
(1)求$△ OAD$的面积;
(2)求点$C'$的坐标.

21. (1)
∵ 点$A$的坐标是$(4,2)$,四边形$ABOC$为长方形,$\therefore AB=OC=4$,$AC=OB=2$,$AB // OC$,$∠ ABO=90°$,$\therefore ∠ BAO=∠ AOC$.由折叠可得$∠ AOC=∠ AOC'$,$\therefore ∠ BAO=∠ AOC'$,$\therefore DO=DA$.设$DO=DA=m$,则$BD=4-m$,$\therefore m^2=2^2+(4-m)^2$,解得$m=\dfrac{5}{2}$,$\therefore △ OAD$的面积为$\dfrac{1}{2}AD · OB=\dfrac{1}{2} × \dfrac{5}{2} × 2=\dfrac{5}{2}$.
(2) 如图,过$C'$作$C'H ⊥ OC$于点$H$,交$AB$于点$M$,$\therefore C'H ⊥ AB$,$MH=AC=2$.由折叠可得$AC=AC'=2$,$∠ AC'O=∠ ACO=90°$,$\therefore C'D=\sqrt{AD^2-C'A^2}=\dfrac{3}{2}$.$\because \dfrac{1}{2}AC' · C'D=\dfrac{1}{2}AD · C'M$,即$\dfrac{1}{2} × 2 × \dfrac{3}{2}=\dfrac{1}{2} × \dfrac{5}{2} · C'M$,$\therefore C'M=\dfrac{6}{5}$,$\therefore C'H=\dfrac{6}{5}+2=\dfrac{16}{5}$,$AM=\sqrt{C'A^2-C'M^2}=\dfrac{8}{5}$,$\therefore BM=4-\dfrac{8}{5}=\dfrac{12}{5}$,$\therefore C'(\dfrac{12}{5},\dfrac{16}{5})$.