**Question:**

Let *ABCD* be a square of side 2*a*. Find the coordinates of the vertices of this square when

(i) A coincides with the origin and *AB* and *AD* are along *OX* and *OY* respectively.

(ii) The centre of the square is at the origin and coordinate axes are parallel to the sides *AB* and *AD* respectively.

**Solution:**

The distance between any two adjacent vertices of a square will always be equal. This distance is nothing but the side of the square.

Here, the side of the square ‘*ABCD*’ is given to be ‘2*a*’.

(i) Since it is given that the vertex ‘*A*’ coincides with the origin we know that the co-ordinates of this point is (0, 0).

We also understand that the side ‘*AB*’ is along the *x*-axis. So, the vertex ‘*B*’ has got to be at a distance of ‘2*a*’ from ‘*A*’.

Hence the vertex ‘*B*’ has the co-ordinates (2*a**, *0).

Also it is said that the side ‘*AD*’ is along the *y*-axis. So, the vertex ‘*D*’ it has got to be at a distance of ‘*2a*’ from ‘*A*’.

Hence the vertex ‘*D*’ has the co-ordinates (0, 2*a*)

Finally we have vertex ‘*C*’ at a distance of ‘2*a*’ both from vertex ‘*B*’ as well as ‘*D*’.

Hence the vertex of ‘*C*’ has the co-ordinates (2*a**, *2*a*)

So, the co-ordinates of the different vertices of the square are

$A(0,0)$

$B(2 a, 0)$

$C(2 a, 2 a)$

$D(0,2 a)$

(ii) Here it is said that the centre of the square is at the origin and that the sides of the square are parallel to the axes.

Moving a distance of half the side of the square in either the ‘*upward*’ or ‘*downward*’ direction and also along either the ‘*right*’ or ‘*left*’ direction will give us all the four vertices of the square.

Half the side of the given square is ‘*a*’.

The centre of the square is the origin and its vertices are (0, 0).* *Moving a distance of ‘*a*’ to the right as well as up will lead us to the vertex ‘*A*’ and it will have vertices (*a, a*).

Moving a distance of ‘*a*’ to the left as well as up will lead us to the vertex ‘*B*’ and it will have vertices (-(−*a**, a*).

Moving a distance of ‘*a*’ to the left as well as down will lead us to the vertex ‘*C*’ and it will have vertices (-(−*a**, *-−*a*).

Moving a distance of ‘*a*’ to the right as well as down will lead us to the vertex ‘*D*’ and it will have vertices (*a,-,−a*).

So, the co-ordinates of the different vertices of the square are

$A(a, a)$

$B(-a, a)$

$C(-a,-a)$

$D(a,-a)$