《2025年名校课堂八年级数学上册沪科版》

【例1】(2023·六安金安区期末节选)如图，直线$y_1 = -\frac{1}{2}x + 1$与直线$y_2 = 2x + 6$分别与$x$轴交于点$A$，$B$，两直线相交于点$P$.
(1)求直线$y_1$，$y_2$与$x$轴的交点坐标；
(2)求三角形$ABP$的面积.

答案：
(1)把y=0代入$y_1 = - \frac{1}{2}x + 1$中，得$- \frac{1}{2}x + 1 = 0$，解得$x = 2$．$\therefore A(2,0)$．把y=0代入$y_2 = 2x + 6$中，得$2x + 6 = 0$，解得$x = - 3$．$\therefore B(-3,0)$．
(2)由
(1)知，$A(2,0)$，$B(-3,0)$，$\therefore AB = 2 - (-3)=5$．解方程$- \frac{1}{2}x + 1 = 2x + 6$，得$x = - 2$．把$x = - 2$代入$y_1 = - \frac{1}{2}x + 1$中，得$y = 2$．$\therefore P(-2,2)$．$\therefore S_{\triangle ABP} = \frac{1}{2} × 5 × 2 = 5$．

1. (2024·合肥45中月考节选)如图，直线$l_1:y = -2x + 4$与$x$轴交于点$B$，$OB = OC$，直线$l_2:y = kx + b$经过点$C$，且与$l_1$交于点$A(1,2)$.
(1)求直线$l_2$的表达式；
(2)记直线$l_2$与$y$轴的交点为$D$，记直线$l_1$与$y$轴的交点为$E$，求三角形$ADE$的面积.

答案：
(1)$\because$直线$l_1$的表达式为$y = - 2x + 4$，当y=0时，x=2．$\therefore B(2,0)$．$\because OB = OC$，$\therefore C(-2,0)$．$\because l_2:y = kx + b$经过点C和点A，$\therefore \begin{cases} -2k + b = 0, \\k + b = 2,\end{cases}$解得$\begin{cases} k = \frac{2}{3}, \\b = \frac{4}{3}.\end{cases}$$\therefore$直线$l_2$的表达式为$y = \frac{2}{3}x + \frac{4}{3}$．
(2)把x=0代入$y = - 2x + 4$中，得y=4．$\therefore E(0,4)$．把x=0代入$y = \frac{2}{3}x + \frac{4}{3}$中，得$y = \frac{4}{3}$．$\therefore D(0,\frac{4}{3})$．$\therefore DE = 4 - \frac{4}{3} = \frac{8}{3}$．$\therefore S_{\triangle ADE} = \frac{1}{2} × \frac{8}{3} × 1 = \frac{4}{3}$．

【例2】(2024·合肥45中期中)如图，在平面直角坐标系中，直线$m:y = -x + b$与直线$n:y = ax + 10(a \neq 0)$交于点$A(-1,5)$，直线$m$，$n$分别与$x$轴交于点$B$，$C$.
(1)求点$B$和点$C$的坐标；
(2)若直线$AC$上存在一点$P$，使得$S_{三角形ABP} = 30$，求点$P$的坐标.

答案：
(1)$\because$直线$m:y = - x + b$与直线$n:y = ax + 10$交于点A(-1,5)，$\therefore \begin{cases} 5 = -(-1) + b, \\5 = -a + 10,\end{cases}$解得$\begin{cases} a = 5, \\b = 4.\end{cases}$$\therefore$直线m的表达式为$y = - x + 4$，直线n的表达式为$y = 5x + 10$．由$0 = - x + 4$，得$x = 4$；由$0 = 5x + 10$，得$x = - 2$．$\therefore B(4,0)$，$C(-2,0)$．
(2)由
(1)可得，$BC = 4 - (-2)=6$．$\therefore S_{\triangle ABC} = \frac{1}{2}BC \cdot y_A = \frac{1}{2} × 6 × 5 = 15$．由
(1)知，直线AC的表达式为$y = 5x + 10$．设$P(p,5p + 10)$．分两种情况讨论：①如图1，当点P在CA的延长线上时，由条件可知，$S_{\triangle BCP} = S_{\triangle ABC} + S_{\triangle ABP} = 15 + 30 = 45$，$\therefore S_{\triangle BCP} = \frac{1}{2}BC \cdot |y_P| = \frac{1}{2} × 6 × (5p + 10) = 45$，解得$p = 1$．$\therefore P(1,15)$；②如图2，当点P在AC的延长线上时，由条件可知，$S_{\triangle BCP} = S_{\triangle ABP} - S_{\triangle ABC} = 15$，$\therefore S_{\triangle BCP} = \frac{1}{2}BC \cdot |y_P| = \frac{1}{2} × 6 × |5p + 10| = 15$，$\therefore |5p + 10| = 5$，解得$p = - 3$或$p = - 1$．当$p = - 1$时，$5p + 10 = 5 ＞ 0$(不符合题意，舍去)，$\therefore P(-3,-5)$．综上所述，点P的坐标为(1,15)或(-3,-5)．