Question

Each side of a square is of length $4$. The centre of the square is $(3,7)$ and one of its diagonals is parallel to $y=x$. Then co-ordinates of its vertices are

• A. $(1,5),(1,9),(5,9),(5,5)$
• B. $(2,5),(2,7),(4,7),(4,4)$
• C. $(2,5),(2,6),(3,5),(3,6)$
• D. $(5,2),(6,2),(5,3),(6,3)$

Answer 1

Answer: (A)

$(1,5),(1,9),(5,9),(5,5)$

From given, we have,

$y-7=1(x-3)$

$y=x+4$

Let $h$ be the point.

$\sqrt{(h-3{)}^2+(h+4-7{)}^2}$

$\frac{4\sqrt{2}}{2}=\sqrt{h^2+9-6h+h^2+9-6h}$

$\frac{32}{4}=2h^2-12h+18$

$4=h^2-6h+9$

$h^2-6h+5=0$

$h=1,5$

Therefore, the vertices are,

$(1,5),(5,5),(5,9),(1,9)$