Given Four friends are seated at, A,B,C and D where A(3,4) B(6,7) C(9,4) and D(6,1)
Now distance=√((x2 - x1)2 + (y2 - y1)2)
AB=√((-)2 + (-)2)
=√
=√2
BC=√((-)2 + (-)2)
=√
=√2
CD=√((-)2 + (-)2)
=√
=√2
AD=√((-)2 + (-)2)
=√
=√2
AC=√((-)2 + (-)2)
=√
=
BD=√((-)2 + (-)2)
=√
=
Hence in ABCD four sides are equal AB=BC=CD=DA=√2
and two digonals are equal AC=BD=units
∴ ABCD forms a square i.e., jarina is correct