《2025年启航新课堂八年级数学上册北师大版》

16. 如图,在平面直角坐标系中,$\triangle ABC顶点的坐标分别为A(2,0)$,$B(0,4)$,$C(-3,2)$.
(1)求$\triangle ABC$的面积；
(2)若点$P的坐标为(m,0)$.
①$AP$的长为______；(用含$m$的式子表示)
②当$S_{\triangle PAB}= 2S_{\triangle ABC}$时,求$m$的值；
(3)若$AC交y轴于点G$,求点$G$的坐标.

答案：

(1)过点$C$作$CD\perp x$轴，垂足为$D$，过点$B$作$BE\perp CD$，交$DC$的延长线于点$E$，过点$A$作$AF\perp BE$，交$EB$的延长线于点$F$。
$\because A(2,0)$，$B(0,4)$，$C(-3,2)$，
$\therefore D(-3,0)$，$E(-3,4)$，$F(2,4)$，$OB = 4$，
$\therefore AD = 5$，$CD = 2$，$BE = 3$，$CE = 2$，$DE = 4$，$BF = 2$，$AF = 4$，
$\therefore S_{\triangle ABC} = S_{长方形ADEF} - S_{\triangle ACD} - S_{\triangle BCE} - S_{\triangle ABF}$
$= AD\cdot DE - \frac{1}{2}AD\cdot CD - \frac{1}{2}CE\cdot BE - \frac{1}{2}BF\cdot AF$
$= 5× 4 - \frac{1}{2}× 5× 2 - \frac{1}{2}× 2× 3 - \frac{1}{2}× 2× 4$
$= 8$。

(2)①$|m - 2|$；
②$\because S_{\triangle PAB} = 2S_{\triangle ABC}$，$\therefore \frac{1}{2}AP\cdot BO = 2× 8$，
$\therefore AP = |m - 2| = 8$，
$\therefore m - 2 = 8$或$m - 2 = -8$，$\therefore m = 10$或$m = -6$。
(3)过点$C$作$CH\perp y$轴，垂足为$H$。
$\because A(2,0)$，$B(0,4)$，$C(-3,2)$，
$\therefore OA = 2$，$OB = 4$，$CH = 3$。
$\because S_{\triangle ABC} = S_{\triangle ABC} + S_{\triangle BCG}$，
$\therefore \frac{1}{2}BG× 2 + \frac{1}{2}BG× 3 = 8$，
$\therefore BG = \frac{16}{5}$，$\therefore OG = OB - BG = 4 - \frac{16}{5} = \frac{4}{5}$，
$\therefore G(0,\frac{4}{5})$。