Question

Examine whether the following points taken in order form a square.

( - 1, 2), (1, 0), (3, 2) and (1, 4)

Answer 1

AB= rad (-1-1)^2+(2-0)^2
AB=rad -2^2+2^2
AB= rad4+4
AB= rad 8
AB= 2 rad 2

BC= rad (1-3)^2+(0-2)^2
BC= rad (-2)^2+(-2)^2
BC= rad 4+4
BC= rad 8
BC=2 rad 2

CD= rad (3-1)^2+(2-4)^2
CD=rad 2^2+(-2)^2
CD= rad 4+4
CD= rad 8
CD=2 rad 2

DA=rad (-1-1)^2 + (2-4)^2
DA= rad (-2)^2+(-2)^2
DA=rad 4+4
DA=rad 8
DA=2rad 2

=> ABDC square

Answer 2

let point A= (-1,2),  B = (1,0),  C = (3,2),    D = (1,4)
distance between AB = √{(1+1)² + (0-2)²} = 2√2
and so for BC = CD = DA = 2√2
distance between AC = √{(3+1)² + (2-2)²} = 4
and so for BD = 4
slope of AB = (2-0)/(-1-1) = -1 = m1
slope of BC = (2-0)/(3-1) = 1 = m2
slope of CD = (4-2)/(1-3) = -1 = m3
slope of DA = (2-4)/(-1-1) = 1 = m4
m1*m2 = m2*m3= m3*m4 = m4*m1 = -1
hence all side are perpendicular ,and
all sides have equal length and both diagonal are equal hence point form a square.