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

Question

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 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Each family has parallel lines having the distance between them as

1/ 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

.

Q. 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

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