AMC12 2005 B

AMC12 2005 B · Q7

AMC12 2005 B · Q7. It mainly tests Area & perimeter, Coordinate geometry.

What is the area enclosed by the graph of $|3x|+|4y|=12$?

由方程 $|3x|+|4y|=12$ 的图象所围成的面积是多少？

(A) 6 6

(B) 12 12

(D) 24 24

(E) 25 25

Answer

Correct choice: (D)

正确答案：（D）

Solution

If we get rid of the absolute values, we are left with the following 4 equations (using the logic that if $|a|=b$, then $a$ is either $b$ or $-b$): \begin{align*} 3x+4y=12 \\ -3x+4y=12 \\ 3x-4y=12 \\ -3x-4y=12 \end{align*} We can then put these equations in slope-intercept form in order to graph them. \begin{align*} 3x+4y=12 \,\implies\, y=-\dfrac{3}{4}x+3\\ -3x+4y=12\,\implies\, y=\dfrac{3}{4}x+3\\ 3x-4y=12\,\implies\, y=\dfrac{3}{4}x-3\\ -3x-4y=12\,\implies\, y=-\dfrac{3}{4}x-3\end{align*} Now you can graph the lines to find the shape of the graph: We can easily see that it is a rhombus with diagonals of $6$ and $8$. The area is $\dfrac{1}{2}\times 6\times8$, or $\boxed{\mathrm{(D)}\ 24}$

如果去掉绝对值，我们得到下面 4 个方程（利用若 $|a|=b$，则 $a$ 可能为 $b$ 或 $-b$）： \begin{align*} 3x+4y=12 \\ -3x+4y=12 \\ 3x-4y=12 \\ -3x-4y=12 \end{align*} 然后将这些方程化为斜截式以便作图。 \begin{align*} 3x+4y=12 \,\implies\, y=-\dfrac{3}{4}x+3\\ -3x+4y=12\,\implies\, y=\dfrac{3}{4}x+3\\ 3x-4y=12\,\implies\, y=\dfrac{3}{4}x-3\\ -3x-4y=12\,\implies\, y=-\dfrac{3}{4}x-3\end{align*} 作出这些直线即可得到图形的形状：容易看出它是一个菱形，两条对角线分别为 $6$ 和 $8$。面积为 $\dfrac{1}{2}\times 6\times8$，即 $\boxed{\mathrm{(D)}\ 24}$

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