《2025年假期总动员年度系统复习八年级数学人教版四川师范大学出版社》

24. 如图，已知直线$y = x + 3$的图象与$x$轴、$y$轴交于$A$，$B$两点，直线$l$经过原点，与线段$AB$交于点$C$，把$\triangle AOB$的面积分为$2:1$的两部分，求直线$l$的解析式。

解:$\because$直线$y=x+3$的图象与$x,y$轴交于$A,B$两点.
$\therefore A$点坐标为$(-3,0)$,$B$点坐标为$(0,3)$.
$\therefore |OA|=3$,$|OB|=3$.
$\therefore S_{△AOB}=\frac {1}{2}|OA|\cdot |OB|=\frac {1}{2}×3×3=\frac {9}{2}$.
设直线$l$的解析式为$y=kx(k≠0)$.
$\because$直线$l$把$△AOB$的面积分为$2:1$,直线$l$与线段$AB$交于点$C$.
$\therefore$分两种情况来讨论:
①当$S_{△AOC}:S_{△BOC}=2:1$时,设$C$点坐标为$(x_{1},y_{1})$.
又$\because S_{△AOB}=S_{△AOC}+S_{△BOC}=\frac {9}{2}$,
$\therefore S_{△AOB}=\frac {9}{2}×\frac {2}{3}=3$.
即$S_{△AOC}=\frac {1}{2}\cdot |OA|\cdot |y_{1}|=\frac {1}{2}×3×|y_{1}|=3$.
$\therefore y_{1}=\pm 2$,由图示可知取$y_{1}=2$.
又$\because$点$C$在直线$AB$上,
$\therefore 2=x_{1}+3$,$\therefore x_{1}=-1$.
$\therefore C$点坐标为$(-1,2)$.
把$C$点坐标$(-1,2)$代入$y=kx$中,得
$2=-1\cdot k$,$\therefore k=-2$.
$\therefore$直线$l$的解析式为$y=-2x$.
②当$S_{△AOC}:S_{△BOC}=1:2$时,设$C$点坐标为$(x_{2},y_{2})$.
又$\because S_{△AOB}=S_{△AOC}+S_{△BOC}=\frac {9}{2}$,
$\therefore S_{△AOC}=\frac {9}{2}×\frac {1}{3}=\frac {3}{2}$,
即$S_{△AOC}=\frac {1}{2}\cdot |OA|\cdot |y_{2}|=\frac {1}{2}\cdot 3\cdot |y_{2}|=\frac {3}{2}$.
$\therefore y_{2}=\pm 1$,由图示可知取$y_{2}=1$.
又$\because$点$C$在直线$AB$上,
$\therefore 1=x_{2}+3$,$\therefore x_{2}=-2$.
把$C$点坐标$(-2,1)$代入$y=kx$中,得
$1=-2k$,$\therefore k=-\frac {1}{2}$.
$\therefore$直线$l$的解析式为$y=-\frac {1}{2}x$.
$\therefore$直线$l$的解析式为$y=-2x$或$y=-\frac {1}{2}x$.

答案： 解:$\because$直线$y=x+3$的图象与$x,y$轴交于$A,B$两点.
$\therefore A$点坐标为$(-3,0)$,$B$点坐标为$(0,3)$.
$\therefore |OA|=3$,$|OB|=3$.
$\therefore S_{△AOB}=\frac {1}{2}|OA|\cdot |OB|=\frac {1}{2}×3×3=\frac {9}{2}$.
设直线$l$的解析式为$y=kx(k≠0)$.
$\because$直线$l$把$△AOB$的面积分为$2:1$,直线$l$与线段$AB$交于点$C$.
$\therefore$分两种情况来讨论:
①当$S_{△AOC}:S_{△BOC}=2:1$时,设$C$点坐标为$(x_{1},y_{1})$.
又$\because S_{△AOB}=S_{△AOC}+S_{△BOC}=\frac {9}{2}$,
$\therefore S_{△AOB}=\frac {9}{2}×\frac {2}{3}=3$.
即$S_{△AOC}=\frac {1}{2}\cdot |OA|\cdot |y_{1}|=\frac {1}{2}×3×|y_{1}|=3$.
$\therefore y_{1}=\pm 2$,由图示可知取$y_{1}=2$.
又$\because$点$C$在直线$AB$上,
$\therefore 2=x_{1}+3$,$\therefore x_{1}=-1$.
$\therefore C$点坐标为$(-1,2)$.
把$C$点坐标$(-1,2)$代入$y=kx$中,得
$2=-1\cdot k$,$\therefore k=-2$.
$\therefore$直线$l$的解析式为$y=-2x$.
②当$S_{△AOC}:S_{△BOC}=1:2$时,设$C$点坐标为$(x_{2},y_{2})$.
又$\because S_{△AOB}=S_{△AOC}+S_{△BOC}=\frac {9}{2}$,
$\therefore S_{△AOC}=\frac {9}{2}×\frac {1}{3}=\frac {3}{2}$,
即$S_{△AOC}=\frac {1}{2}\cdot |OA|\cdot |y_{2}|=\frac {1}{2}\cdot 3\cdot |y_{2}|=\frac {3}{2}$.
$\therefore y_{2}=\pm 1$,由图示可知取$y_{2}=1$.
又$\because$点$C$在直线$AB$上,
$\therefore 1=x_{2}+3$,$\therefore x_{2}=-2$.
把$C$点坐标$(-2,1)$代入$y=kx$中,得
$1=-2k$,$\therefore k=-\frac {1}{2}$.
$\therefore$直线$l$的解析式为$y=-\frac {1}{2}x$.
$\therefore$直线$l$的解析式为$y=-2x$或$y=-\frac {1}{2}x$.