# Differing by $2$ “puzzle”

I've been thinking about the following:

For (1) I drew $y= 0$, $y=2x$ and $y = 2x$, $y = 4x$. The expression I wrote down if $y = mx + b$ is one of the lines then $y_2 = (m \pm 2)x + b$ is the other.

For (2) I solved $y= mx + b$ at $y=0$ for $x$. Then we get $x_2 = \frac{-b}{m} \pm 2$ for the new $x$-intercepts and hence two new lines $y_2 = mx + b \pm 2m$. I drew $y = x$ and $y= x \pm 2$.

For (3) we get two new lines $y_2 = mx + b \pm 2$. Again I drew $y = x$ and $y= x \pm 2$.

Now (4) is where I am missing something. I combined (1),(2) and (3) to get lines that are supposed to differ by $2$ in slope, $x$- and $y$-intercept, $y_2 = (m+2)x + b + 2 + 2m$ (if all differ by $+2$), for example:

$y_1 = x + 2, y_2 = 3x + 6$. But this is wrong since the $y$-intercept now differs by $4$.

Question 1: How do I answer (4)?

Question 2: What is special about $2$? I don't understand what the puzzle is getting at. It appears to me that I could replace $2$ by $n$ and do the exact same questions.

Thank you!

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## 2 Answers

For the second, third, and fourth parts, it's convenient to remember the intercept-intercept form of a line (not passing through the origin):

$$\frac{x}{a} + \frac{y}{b} = 1$$

where $a\ne0$ is the $x$-intercept, and $b\ne 0$ is the $y$-intercept, and $-b/a$ is the slope. So, with the above as the equation for one of the lines in question, we can easily answer the puzzle's second and third parts with these equations $$(2) \quad \frac{x}{a\pm2} + \frac{y}{b} = 1 \qquad \qquad (3) \quad \frac{x}{a}+\frac{y}{b\pm2}=1$$

with appropriate restrictions on $a$ and $b$. (Note that these fix the "other" intercept and allow the slope to vary, rather than keeping the slope fixed.) For the fourth part, you'd adjust the intercepts, but also relate the corresponding slopes:

$$\frac{x}{a\pm 2} + \frac{y}{b\pm 2} = 1 \qquad \mathrm{where} \qquad -\frac{b\pm2}{a\pm2} = -\frac{b}{a} \pm 2$$

with all "$\pm$"s independent. Here, $a$ and $b$ are dependent, and we can solve for one in terms of the other. To deal with the rampant "$\pm$"s, let's write

$$-\frac{b+2q}{a+2p} = - \frac{b}{a} + 2r$$ with $p,q,r \in \{1,-1\}$. Solving for $b$ gives $$b = a ( a pr + 2 r + q )$$ so that the lines are $$\frac{x}{a}+ \frac{y}{a(apr+2r+q)} = 1 \qquad \mathrm{and} \qquad \frac{x}{a+2p} + \frac{y}{a(apr+2r+q)+2q} = 1$$

I don't see anything particularly special about "$2$" in the above, though.

Now, if one of the lines passes through the origin, we have

$$y = m x$$

so that its counterpart for part 4 must have an equation of the form

$$\frac{x}{\pm 2} + \frac{y}{\pm 2} = 1 \qquad \mathrm{with} \qquad - \frac{\pm 2}{\pm 2} = m \pm 2$$

Thus, $m \in \{1,-3,3,-1\}$. It's perhaps an interesting coincidence that these are the coefficients of $(1-z)^3$, but replacing $2$ with, say, $k\ne 0$ simply gives $m = \pm k \pm 1$.

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You could consider the case thatthe first line has $x$-intercept and $y$-intercept equal to $0$ and the other has $x$-intercept and $y$-intercept equal to $2$. Then the slope of the second line is $-1$, hence the slope of the first line should be $+1$ or $-3$. There is nothing special about the difference $2$. For any $c>0$, the lines $y=c-x$ and $y=(-1-c)x$ have $x$-intercepts, $y$-intercepts and slopes differing by $c$. As long as $c\ne1$, the second line may be replaced by $y=(c-1)x$. At any rate, it is also allowed to translate both lines by the same amount vertically.

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If you translate them vertically, the distance between the $x$-intercepts will change. – Brian M. Scott Oct 14 '12 at 11:10
Thank you! How did you reach that expression for (4)? – Rudy the Reindeer Oct 14 '12 at 11:50