Question

Show that the following points are the vertices of a square:

(i) A(3, 2), B (0, 5), C(-3, 2) and D(0, -1)
(ii) A(6, 2), B(2, 1), C(1, 5) and D(5, 6)
(iii) A(0, -2), B(3, 1), C(0, 4) and D(-3, 1)

Answer 1

(i)The given points are A (3,2), B(0,5) and C(-3,2), D(0,-1). Then

AB =$\sqrt{(0-3{)}^2+(5-2{)}^2}$

= $\sqrt{(-3{)}^2+(3{)}^2}$

=$\sqrt{9+9}$

= $\sqrt{18}$

=$3\sqrt{2}$ units

BC =$\sqrt{(-3-0{)}^2+(2-5{)}^2}$

= $\sqrt{(-3{)}^2+(-3{)}^2}$

= $\sqrt{9+9}$

= $\sqrt{1}8$

=$3\sqrt{2}$ units

CD = $\sqrt{(0+3{)}^2+(-1-2{)}^2}$

= $\sqrt{(3{)}^2+(-3{)}^2}$

= $\sqrt{9+9}$

=$\sqrt{18}$

=$3\sqrt{2}$ units

DA = $\sqrt{(0-3{)}^2+(-1-2{)}^2}$

=$\sqrt{(-3{)}^2+(-3{)}^2}$

= $\sqrt{9+9}$

=$\sqrt{18}$

=$3\sqrt{2}$ units

Therefore AB = BC = CD = DA =$3\sqrt{2}$ units

Also,

AC $=\sqrt{(-3-3{)}^2+(2-2{)}^2}$

$=\sqrt{(-6{)}^2+(0{)}^2}$

$=\sqrt{36}$

= 6 units

BD $=\sqrt{(0-0{)}^2+(-1-5{)}^2}$

$=\sqrt{(0{)}^2+(-6{)}^2}$

$=\sqrt{36}$

= 6 units

Thus, diagonal AC = diagonal BD

Therefore, the given points form a square.

(ii) A(6, 2), B(2, 1), C(1, 5) and D(5, 6)

Solution

The given points are A (6,2), B(2,1) and C(1,5), D(5,6). Then

$AB=\sqrt{(2-6{)}^2+(1-2{)}^2}$

$=\sqrt{(-4{)}^2+(-1{)}^2}$

$=\sqrt{16+1}$

$=\sqrt{17}$ units

$BC=\sqrt{(1-2{)}^2+(5-1{)}^2}$

$=\sqrt{(-1{)}^2+(-4{)}^2}$

$=\sqrt{1+16}$

$=\sqrt{17}$ units

$CD=\sqrt{(5-1{)}^2+(6-5{)}^2}$

$=\sqrt{(4{)}^2+(1{)}^2}$

$=\sqrt{16+1}$

$=\sqrt{17}$units

$DA=\sqrt{(5-6{)}^2+(6-2{)}^2}$

$=\sqrt{(1{)}^2+(4{)}^2}$

$=\sqrt{1+16}$

$=\sqrt{17}$ units

Therefore $AB=BC=CD=DA=\sqrt{17}$ units

Also,
$AC=\sqrt{(1-6{)}^2+(5-2{)}^2}$

$=\sqrt{(-5{)}^2+(3{)}^2}$

$=\sqrt{25+9}$

$=\sqrt{34}$units

$BD=\sqrt{(5-2{)}^2+(6-1{)}^2}$

$=\sqrt{(3{)}^2+(5{)}^2}$

$=\sqrt{9+25}$

$=\sqrt{34}$ units

Thus, diagonal AC = diagonal BD

Therefore, the given points form a square.

(iii)A(0, -2), B ( 3, 1), C (0, 4) and D (-3, 1)

Solution

The given points are P (0,-2), Q(3,1) and R(0,4), S(-3,1). Then

$PQ=\sqrt{(3-0{)}^2+(1+2{)}^2}$

$=\sqrt{(3{)}^2+(3{)}^2}$

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$units

$QR=\sqrt{(0-3{)}^2+(4-1{)}^2}$

$=\sqrt{(-3{)}^2+(3{)}^2}$

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$units

$RS=\sqrt{(-3-0{)}^2+(1-4{)}^2}$

$=\sqrt{(-3{)}^2+(-3{)}^2}$

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$units

$SP=\sqrt{(-3-0{)}^2+(1+2{)}^2}$

$=\sqrt{(-3{)}^2+(3{)}^2}$

=$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$units

Therefore PQ = QR = RS = SP $=3\sqrt{2}$units

Also,
$PR=\sqrt{(0-0{)}^2+(4+2{)}^2}$

$=\sqrt{(0{)}^2+(6{)}^2}$

$=\sqrt{36}$

= 6 units

$QS=\sqrt{(-3-3{)}^2+(1-1{)}^2}$

$=\sqrt{(-6{)}^2+(0{)}^2}$

$=\sqrt{36}$

= 6 units

Thus, diagonal PR = diagonal QS

Therefore, the given points form a square.