Consider the fourteen lines in the plane given by $y=x+r, \quad y=-x+r, \quad$ where $r \in\{0,1,2,3,4,5…
Consider the fourteen lines in the plane given by $y=x+r, \quad y=-x+r, \quad$ where $r \in\{0,1,2,3,4,5,6\}$. The number of squares formed by these lines, whose sides are of length $\sqrt{2}$, is :
9
16
25
36
Solution
We have,
$y=x+r$ $\ldots$ (i)
and $y=-x+r$ $\ldots$ (ii)
The given lines are perpendicular to each other.
perpendicular distance
$=\frac{\left|r_1-r_2\right|}{\sqrt{2}}=\sqrt{2}$
$\Rightarrow \quad r_1-r_2=2$
$\Rightarrow$ The difference between the $y$-intercepts $=2$
This can happen for five combinations $\{(0,2),(1,3),(2,4),(3,5),(4,6)\}$
The difference between the $x$-intercepts $=2$
This can happen for five combination the total number of squares $=5 \times 5=25$