The graph of the linear function $t$ has intercepts at $(x,0)$ and $(0,y)$. If $x \neq y$ and $x+y=0$, is the slope negative or positive?

Question

“The graph of the linear function $t$ has intercepts at $(x,0)$ and $(0,y)$ in the $xy$-plane. If $x\neq y$ and $x + y = 0$, which of the following is true about the slope of the graph of $t$?”

I was having a discussion with a student who insisted the answer to this question is that it has a positive slope. Another colleague and myself both found the slope for this to be negative and I cannot understand how the student finds the slope to be positive. Is the student correct in a way that I cannot see? Is the slope negative or positive?

Answer 1

The fact that $x+y=0$ tells you that $x$ and $y$ have opposite signs. This tells you that the slope cannot be negative.

Since $x \neq y$ the line can't go through the origin, so no monkey business there. The slope can't be zero, or infinite, because then one of $x$ or $y$ would be undefined.

So you can say that the slope is defined, and strictly positive.

Answer 2

Since $x\not=y$ and $x+y=0$, there are only two possible options for the intercepts. Let's assume that $x=a$ is positive and $y=-a$ is negative. The two intercepts will be $(a,0)$ and $(0,-a)$, and the slope of a line between those points is $\frac{a-0}{0-(-a)}=\frac{a}{a}=1$ and therefore has positive slope.

Likewise, assume that $x=-a$ is negative and $y=a$ is positive. The two intercepts will be $(-a,0)$ and $(0,a)$, and the slope of a line through those points will be $\frac{-a-0}{0-a}=\frac{-a}{-a}=1$, and also has positive slope.

As a result, all possible graphs will have positive slopes.