Question

直线 $\ell_1$ 的方程为 $3x - 2y = 1$ ，并经过点 $A = (-1, -2)$ 。直线 $\ell_2$ 的方程为 $y = 1$ ，与 $\ell_1$ 交于点 $B$ 。直线 $\ell_3$ 斜率为正，经过点 $A$ ，并与 $\ell_2$ 交于点 $C$ 。 $\triangle ABC$ 的面积为 $3$ 。 $\ell_3$ 的斜率是多少？

Line $\ell_1$ has equation $3x - 2y = 1$ and goes through $A = (-1, -2).$ Line $\ell_2$ has equation $y = 1$ and meets line $\ell_1$ at point $B.$ Line $\ell_3$ has positive slope, goes through point $A,$ and meets $\ell_2$ at point $C.$ The area of $\triangle ABC$ is $3.$ What is the slope of $\ell_3?$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$1$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

答案：B

知识点：坐标几何三角形面积斜率

难度评级：1460

解答：

联立 $3x - 2y = 1$ 与 $y = 1$ ，得 $B = (1, 1)$ 。点 $A = (-1, -2)$ 到直线 $y = 1$ 的距离为 $3$ ，所以 $\tfrac12\cdot BC\cdot 3 = 3$ ，从而 $BC = 2$ 。于是 $C = (3, 1)$ 或 $C = (-1, 1)$ ；后者会使 $\ell_3$ 竖直，所以 $C = (3, 1)$ 。因此斜率为 $\dfrac{1 - (-2)}{3 - (-1)} = \dfrac34$ 。正确答案是 B。

Solving $3x - 2y = 1$ with $y = 1$ gives $B = (1, 1).$ The distance from $A = (-1, -2)$ to the line $y = 1$ is $3,$ so $\tfrac12\cdot BC\cdot 3 = 3$ gives $BC = 2.$ Then $C = (3, 1)$ or $C = (-1, 1);$ the latter makes $\ell_3$ vertical, so $C = (3, 1)$ and the slope is $\dfrac{1 - (-2)}{3 - (-1)} = \dfrac34.$ Thus, the correct answer is B.

2013 AMC 12B 第 8 题

答案：B

难度评级：1460

解答：

其他年份的第 8 题