1. (2024·唐山期末)在平面直角坐标系中,以方程$2x - 3y = 6$的解为坐标的点组成的图形是 ()

B

2. (2025·西安校级月考)在同一平面直角坐标系中,直线$y = -x + 4与y = 2x + m相交于点P(3,n)$,则关于$x,y的方程组\begin{cases}x + y - 4 = 0,\\2x - y + m = 0\end{cases}$的解为 ()
A. $\begin{cases}x = -1,\\y = 5\end{cases}$
B. $\begin{cases}x = 1,\\y = 3\end{cases}$
C. $\begin{cases}x = 3,\\y = 1\end{cases}$
D. $\begin{cases}x = 9,\\y = -5\end{cases}$

C

3. (呼和浩特中考改编)若以二元一次方程$x + 2y - b = 0的解为坐标的点(x,y)都在直线y = -\frac{1}{2}x + b - 1$上,则常数$b = $______.

2

4. (1)(甘孜州中考改编)已知一次函数$y = kx + 3和y = -x + b的图象交于点P(2,4)$,则$\begin{cases}x = 2,\\y = 4\end{cases}$是方程组______的解.
(2)已知两条直线$y = mx - 1与y = 3x + 1$平行,则$m = $______,此时方程组$\begin{cases}mx - y = 1,\\3x - y = -1\end{cases}$的解的情况为______.

(1)$\begin{cases}x - 2y = -6, \\x + y = 6\end{cases}$
(2)3无解

5. 如图,过点$A(-2,0)的直线l_1:y = kx + b与直线l_2:y = -x + 1交于点P(-1,a)$.
(1)求直线$l_1$对应的表达式;
(2)直接写出方程组$\begin{cases}y = kx + b,\\y = -x + 1\end{cases}$的解;
(3)求四边形$PAOC$的面积.

(1)把P(−1,a)代入y=−x+1,得a=2,则点P坐标为(−1,2).把A(−2,0),P(−1,2)代入y=kx+b,得$\begin{cases}0 = -2k + b, \\2 = -k + b,\end{cases}$解得$\begin{cases}k = 2, \\b = 4,\end{cases}$∴直线$l_1$的表达式为y=2x+4.
(2)方程组$\begin{cases}y = kx + b, \\y = -x + 1\end{cases}$的解为$\begin{cases}x = -1, \\y = 2.\end{cases}$
(3)∵y=−x+1交x轴于点B,交y轴于点C,∴B(1,0),C(0,1),∴四边形PAOC的面积=S△ABP−S△BOC=$\frac{1}{2}$×3×2−$\frac{1}{2}$×1×1=$\frac{5}{2}$.

6. 方程组$\begin{cases}ax + y = 3,\\x + by = -1\end{cases}$所对应的一次函数图象如图所示,则$2a + b$的值为 ()

A. -5
B. 3
C. 5
D. -3

A　解析:根据题意，得方程组$\begin{cases}ax + y = 3, \\x + by = -1\end{cases}$的解为$\begin{cases}x = -2, \\y = -1,\end{cases}$所以$\begin{cases}-2a - 1 = 3, \\-2 - b = -1,\end{cases}$解得$\begin{cases}a = -2, \\b = -1,\end{cases}$所以2a+b=2×(−2)−1=−5.故选A.

7. (2024·福建期末)如图,过第一象限上$A点的直线是方程x - y = b(b ＜ -1)$的图象,若点$A的坐标恰为关于x,y的二元一次方程组\begin{cases}x - y = b,\\ax - y = 1\end{cases}$的解,则$a$的值可能是 ()

A. -1
B. 0
C. 1
D. 2

D　解析:∵点A在第一象限,∴x＞0,且$\begin{cases}x - y = b, \\ax - y = 1,\end{cases}$② - ①得(a - 1)x=−(b - 1),∴x=$\frac{−(b - 1)}{a - 1}$＞0.∵b＜−1,∴−(b - 1)＞0,∴a - 1＞0,∴a＞1.故选D.