2017 AMC 10A Problem 12

Let us case on which of the values are equal. If $3 = x + 2,$ then $x = 1.$ This also tells us that $$y - 4 \leq 3$$$$y \leq 7.$$ This describes a ray starting at $(1, 7)$ and extending in the negative $y$ direction. Similarly, if $3 = y - 4,$ then $y = 7$ and $$x + 2 \leq 3$$$$x \leq 1.$$ This also describes a ray starting at $(1, 7)$ but instead extending in the negative $x$ direction. Finally, if $x + 2 = y - 4,$ then we have the line $y = x + 6.$ Furthermore, we have that $$3 \leq x + 2$$$$x \geq 1$$ and $$3 \leq y - 4$$$$y \geq 7.$$ Note that if one of these conditions is met, the other is also necessarily true due to the equation of the line. If $y = 7,$ then $x = 1.$ The other points are along the line, where $y \gt 7$ and $x \gt 1.$ This describes another ray that starts at $(1, 7)$ and goes off in some third direction. All three cases result in rays originating from $(1, 7)$ that all go in different directions. Thus, E is the correct answer.