AMC12 2020 A

AMC12 2020 A · Q12

AMC12 2020 A · Q12. It mainly tests Manipulating equations, Coordinate geometry.

Line $\ell$ in the coordinate plane has equation $3x-5y+40=0$. This line is rotated $45^\circ$ counterclockwise about the point $(20,20)$ to obtain line $k$. What is the $x$-coordinate of the $x$-intercept of line $k$?

坐标平面中的直线 $\ell$ 的方程为 $3x-5y+40=0$。这条直线绕点 $(20,20)$ 逆时针旋转 $45^\circ$ 得到直线 $k$。直线 $k$ 的 $x$ 截距的 $x$ 坐标是多少？

(A) 10 10

(B) 15 15

(D) 25 25

(E) 30 30

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Note that the point (20, 20) lies on both line $\ell$ and line $k$. The slope of a nonhorizontal, nonvertical line is the tangent of the angle it makes in the upper half-plane with the rightward-pointing $x$-axis. Because line $\ell$ has slope $\frac{3}{5}$, it follows that the slope of line $k$ will be $$ \tan\left(\arctan\frac{3}{5}+45^\circ\right)=\frac{\frac{3}{5}+1}{1-\frac{3}{5}\cdot 1}=4. $$ Therefore an equation of line $k$ is $y-20=4(x-20)$, and this line has $x$-intercept (15, 0).

答案（B）：注意点 (20, 20) 同时在直线 $\ell$ 和直线 $k$ 上。非水平、非竖直直线的斜率等于它在上半平面内与向右的 $x$ 轴所成角的正切。由于直线 $\ell$ 的斜率为 $\frac{3}{5}$，可得直线 $k$ 的斜率为 $$ \tan\left(\arctan\frac{3}{5}+45^\circ\right)=\frac{\frac{3}{5}+1}{1-\frac{3}{5}\cdot 1}=4. $$ 因此，直线 $k$ 的一个方程是 $y-20=4(x-20)$，并且这条直线的 $x$ 轴截距为 (15, 0)。

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