1969 AMC 12 第 18 题

Question

下列两个图形

(x-y+2)(3x+y-4)=0

与

(x+y-2)(2x-5y+7)=0

的公共点个数为：

The number of points common to the graphs of

(x-y+2)(3x+y-4)=0

and

(x+y-2)(2x-5y+7)=0

is:

2

4

6

16

无穷多个

infinite

每个因式分解后的方程都表示两条直线

Each factored equation represents a pair of lines

将第一组中的每条直线分别与第二组中的每条直线求交，并检查交点是否互异

Intersect each line from the first pair with each line from the second pair and check distinctness

第一个图形由直线

x-y+2=0

、

3x+y-4=0

组成，第二个图形由直线

x+y-2=0

、

2x-5y+7=0

组成。第一组中的每条直线都与第二组中的每条直线相交。解这四组方程，得到四个互异的点：

\begin{gathered} (0,2),\quad(-1,1),\\ (1,1),\quad\left(\frac{13}{17},\frac{29}{17}\right) \end{gathered}\text{。}

因而共有

4

个公共点。

所以正确答案是 B。

The first graph is the pair

x-y+2=0,

and

3x+y-4=0,

and the second is

x+y-2=0,

and

2x-5y+7=0.

Each line in the first pair meets each line in the second pair. Solving the four pairings gives four distinct points:

\begin{gathered} (0,2),\quad(-1,1),\\ (1,1),\quad\left(\frac{13}{17},\frac{29}{17}\right). \end{gathered}

Thus there are

4

common points.

Therefore, the correct answer is B.

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