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  4. In a plane there are two families of lines: y=x+r, y =-x+r, where r ∈ 0,1,2,3,4 . The number of the squares of the diagonal of length 2 formed by these lines is

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Q. In a plane there are two families of lines: $y=x+r, y$ $=-x+r$, where $r \in\{0,1,2,3,4\}$. The number of the squares of the diagonal of length 2 formed by these lines is

Straight Lines

Solution:

Each family has parallel lines having the distance between them as $1 / \sqrt{2}$ unit. Both the families are perpendicular to each other. So, to form a square of diagonal 2 units, lines of alternate pair are to be chosen.
Both the families have three such pairs. So, the number of squares possible is $3 \times 3=9$.