【例4】如图,直线$l_1:y= 2x+3与l_2:y= kx-1交于点P$,且点$P的横坐标为-1$,已知直线$l_1$,$l_2分别交y轴于A$,$B$两点.
(1)求$k$的值;
(2)已知$C是x$轴上一点,若$\triangle APC的面积与\triangle ABP$的面积相等,求点$C$的坐标.
[思路导引](1)先求出点$P$的坐标,再代入$y= kx-1$中,求出$k$的值即可;
(2)先求出$S_{\triangle ABP}= 2$,再设点$C的坐标为(x,0)$,设$l_1与x轴的交点为E$,利用$S_{\triangle APC}= S_{\triangle AEC}-S_{\triangle PEC}建立方程求出x$的值,即可得到点$C$的坐标.

[答案]解:(1)$\because直线l_1与l_2交于点P$,且点$P的横坐标为-1$,
$\therefore令x= -1$,则$y= 2×(-1)+3= 1$,$\therefore点P的坐标为(-1,1)$.
把$P(-1,1)代入y= kx-1$,得$-k-1= 1$,解得$k= -2$.
(2)$\because直线l_1:y= 2x+3与y轴交于点A$,$\therefore令x= 0$,则$y= 2×0+3= 3$,
$\therefore点A的坐标为(0,3)$.
$\because直线l_2:y= -2x-1与y轴交于点B$,$\therefore令x= 0$,则$y= -1$,$\therefore B(0,-1)$,$\therefore AB= 3-(-1)= 4$.
又$\because P(-1,1)$,$\therefore S_{\triangle ABP}= \frac{1}{2}×4×1= 2$.
设$l_1与x轴的交点为E$(图略).
令$y= 0$,则$0= 2x+3$,解得$x= -\frac{3}{2}$,$\therefore点E的坐标为(-\frac{3}{2},0)$.
设点$C的坐标为(x,0)$,则$S_{\triangle APC}= S_{\triangle AEC}-S_{\triangle PEC}= \frac{1}{2}\cdot|x+\frac{3}{2}|\cdot(3-1)= 2$,
解得$x= \frac{1}{2}或x= -\frac{7}{2}$,$\therefore点C的坐标为(\frac{1}{2},0)或(-\frac{7}{2},0)$.

【练4】如图,在平面直角坐标系中,一次函数$y= -\frac{1}{2}x+n的图象与x$轴、$y轴分别交于A$,$B$两点,正比例函数$y= kx的图象与一次函数的图象交于点C(2,4)$.
(1)求$n$,$k$的值;$n=$,$k=$
(2)已知$D是一次函数y= -\frac{1}{2}x+n$的图象上的一个动点,连接$OD$,当$\triangle AOD的面积是\triangle BOC的面积的2$倍时,求点$D$的坐标.点$D$的坐标为

练4解:(1)将C(2,4)代入y = -$\frac{1}{2}$x + n和y = kx中,
得4 = -$\frac{1}{2}$×2 + n,4 = 2k,
解得n = 5,k = 2.
(2)在y = -$\frac{1}{2}$x + 5中,
令x = 0,则y = 5;令y = 0,则0 = -$\frac{1}{2}$x + 5,解得x = 10,
∴A(10,0),B(0,5).
∵C(2,4),
∴$S_{\triangle BOC} = \frac{1}{2} × 5 × 2 = 5$.
设D(t,-$\frac{1}{2}$t + 5).
∵△AOD的面积是△BOC的面积的2倍,
∴$S_{\triangle AOD} = 10$,
∴$S_{\triangle AOD} = \frac{1}{2} × 10 × |-\frac{1}{2}t + 5| = 10$,
解得t = 6或t = 14.
当t = 6时, -$\frac{1}{2}$t + 5 = 2;
当t = 14时, -$\frac{1}{2}$t + 5 = -2,
∴点D的坐标为(6,2)或(14,-2).