Step-by-step explanation:

Let B(x,y) and D(x

1

.y

1

) ids the other two vertices.

In Square ABCD

AB=BC=CD=DA

Hence AB=BC

⇒

(x+1)

2

+(y−2)

2

=

(3−x)

2

+(2−y)

2

[by distance formula]

Squaring both sides

⇒(x+1)

2

+(y−2)

2

=(3−x)

2

+(2−y)

2

⇒x

2

+2x+1+y

2

+4−4y=9+x

2

−6x+4+y

2

−4y

⇒2x+5=13−6x

⇒2x+6x=13−5

⇒8x=8

⇒x=1

In △ABC,∠B=90

∘

[all angles of square are 90

∘

]

Then according to the Pythagorean theorem

AB

2

+BC

2

=AC

2

As AB=BC

∴2AB

2

=AC

2

⇒2(

x+1)

2

+(y−2)

2

)

2

=(

(3−(−1))

2

+(2−2)

2

)

2

⇒2((x+1)

2

+(y−2)

2

)=(3+1)

2

+(2−2)

2

⇒2(x

2

+2x+1+y

2

+4−4y)=(4)

2

Put the value of x=1

⇒2(1

1

+2×1+1+y

2

+4−4y)=16

⇒2(y

2

−4y+8)=16

⇒2y

2

−8y+16=16

⇒2y

2

−8y=0

⇒2y(y−4)=0

⇒(y−4)=0

Hence y=0 or 4.

As diagonals of a square are equal in length and bisect each other at 90

∘

Let P is the midpoint of AC

∴CO-ordinates of P=(

2

3−1

,

2

2+2

)=(1,2)

P is also the midpoint of BD

then co-ordinates of mid-point of BD=co-ordinates of P

⇒(x

1

,y

1

)=1,2

∴x

1

=1,y

1

=2

Then other two vertices of square ABCD are (1,0) or (1,4) and (1,2).