《2025年大联考单元期末测试卷八年级数学上册沪科版》

15. 已知直线$ l $与直线$ y = 2x + 1 $的交点的横坐标是$ 2 $,与直线$ y = x - 8 $的交点的纵坐标是$-7$,求直线$ l $的函数表达式.

答案： 15.解：把$x = 2$代入$y = 2x + 1$，得$y = 5$。把$y = - 7$代入$y = x - 8$，得$x = 1$，$\therefore$直线$l$过点$(2,5)$和点$(1, - 7)$。(4分)
设直线$l$的函数表达式为$y = kx + b$。把$(2,5)$和$(1, - 7)$代入，得$\begin{cases}2k + b = 5,\\k + b = - 7,\end{cases}$
解得$\begin{cases}k = 12,\\b = - 19,\end{cases}\therefore y = 12x - 19$，即直线$l$的函数表达式为$y = 12x - 19$。(8分)

16. 如图,$ CD $是$ \triangle ABC $的角平分线,$ \angle 1 = \angle 2 $,$ \angle B = 30^{\circ} $,求$ \angle ACB $的度数.

答案： 16.解：设$\angle ACB = 2\alpha$。
$\because CD$是$\triangle ABC$的角平分线，$\therefore\angle BCD = \angle ACD = \frac{1}{2}\angle ACB = \alpha$，
$\therefore\angle 1 = \angle B + \angle BCD = 30^{\circ} + \alpha$。(4分)
$\because\angle 1 = \angle 2$，$\therefore\angle 2 = 30^{\circ} + \alpha$。
$\because\angle 2 + \angle B + \angle ACB = 180^{\circ}$，$\therefore 30^{\circ} + \alpha + 30^{\circ} + 2\alpha = 180^{\circ}$，
$\therefore\alpha = 40^{\circ}$，$\therefore\angle ACB = 2\alpha = 80^{\circ}$。(8分)