AMC12 2004 A

AMC12 2004 A · Q12

AMC12 2004 A · Q12. It mainly tests Linear equations, Coordinate geometry.

Let $A = (0,9)$ and $B = (0,12)$. Points $A'$ and $B'$ are on the line $y = x$, and $\overline{AA'}$ and $\overline{BB'}$ intersect at $C = (2,8)$. What is the length of $\overline{A'B'}$?

设 $A = (0,9)$ 和 $B = (0,12)$。点 $A'$ 和 $B'$ 在直线 $y = x$ 上，且 $\overline{AA'}$ 和 $\overline{BB'}$ 相交于 $C = (2,8)$。$\overline{A'B'}$ 的长度是多少？

(A) 2 2

(B) $2\sqrt{2}$ $2\sqrt{2}$

(D) $2 + \sqrt{2}$ $2 + \sqrt{2}$

(E) $3\sqrt{2}$ $3\sqrt{2}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Error creating thumbnail: Unable to save thumbnail to destination The equation of $\overline{AA'}$ can be found using points $A, C$ to be $y - 9 = \left(\frac{9-8}{0-2}\right)(x - 0) \Longrightarrow y = -\frac{1}{2}x + 9$. Similarily, $\overline{BB'}$ has the equation $y - 12 = \left(\frac{12-8}{0-2}\right)(x-0) \Longrightarrow y = -2x + 12$. These two equations intersect the line $y=x$ at $(6,6)$ and $(4,4)$. Using the distance formula or $45-45-90$ right triangles, the answer is $2\sqrt{2}\ \mathrm{(B)}$.

线段 $\overline{AA'}$ 的方程可用点 $A, C$ 求得：$y - 9 = \left(\frac{9-8}{0-2}\right)(x - 0) \Longrightarrow y = -\frac{1}{2}x + 9$。类似地，$\overline{BB'}$ 的方程为 $y - 12 = \left(\frac{12-8}{0-2}\right)(x-0) \Longrightarrow y = -2x + 12$。这两条直线与 $y=x$ 的交点分别为 $(6,6)$ 和 $(4,4)$。用距离公式或 $45-45-90$ 直角三角形可得答案为 $2\sqrt{2}\ \mathrm{(B)}$。

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