1963 AMC 12 第 7 题

Question

给出以下四个方程：

\begin{aligned} (1)\quad&3y-2x=12,\\ (2)\quad&-2x-3y=10,\\ (3)\quad&3y+2x=12,\\ (4)\quad&2y+3x=10 \end{aligned}\text{。}

其中表示两条垂直直线的一组是：

Given the four equations:

\begin{aligned} (1)\quad&3y-2x=12,\\ (2)\quad&-2x-3y=10,\\ (3)\quad&3y+2x=12,\\ (4)\quad&2y+3x=10. \end{aligned}

The pair representing perpendicular lines is:

(1)

和

(4)

(1)

and

(4)

(1)

和

(3)

(1)

and

(3)

(1)

和

(2)

(1)

and

(2)

(2)

和

(4)

(2)

and

(4)

(2)

和

(3)

(2)

and

(3)

把每个方程改写成斜截式

Rewrite each equation in slope-intercept form

两条非竖直直线垂直时，斜率之积为

-1

Perpendicular nonvertical lines have slopes whose product is

-1

直线

(1)

的斜率是

\frac{2}{3}

，直线

(4)

的斜率是

-\frac{3}{2}

。它们的积为

-1

，所以这两条直线互相垂直。

所以正确答案是 A。

Line

(1)

has slope

\frac{2}{3},

and line

(4)

has slope

-\frac{3}{2}.

Their product is

-1,

so these two lines are perpendicular.

Thus, the correct answer is A.

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