AMC12 2016 A

AMC12 2016 A · Q7

AMC12 2016 A · Q7. It mainly tests Manipulating equations, Transformations.

Which of these describes the graph of $x^2(x+y+1)=y^2(x+y+1)$?

以下哪一项描述了方程 $x^2(x+y+1)=y^2(x+y+1)$ 的图像？

(A) two parallel lines 两条平行线

(B) two intersecting lines 两条相交直线

(D) three lines that do not all pass through a common point 三条不都通过一个公共点的直线

(E) a line and a parabola 一条直线和一条抛物线

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The given equation is equivalent to $(x^2-y^2)(x+y+1)=0$, which is in turn equivalent to $(x+y)(x-y)(x+y+1)=0$. A product is 0 if and only if one of the factors is 0, so the graph is the union of the graphs of $x+y=0$, $x-y=0$, and $x+y+1=0$. These are three straight lines, two of which intersect at the origin and the third of which does not pass through the origin. Therefore the graph consists of three lines that do not all pass through a common point.

答案（D）：所给方程等价于 $(x^2-y^2)(x+y+1)=0$，这又等价于 $(x+y)(x-y)(x+y+1)=0$。当且仅当其中一个因子为 0 时，乘积才为 0，因此其图像是 $x+y=0$、$x-y=0$ 和 $x+y+1=0$ 这三条直线图像的并集。这是三条直线，其中两条在原点相交，第三条不经过原点。因此，该图像由三条不全经过同一点的直线组成。

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