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

Solution:

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

$A B=\sqrt{(0-3)^{2}+(5-2)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$ units

$B C=\sqrt{(-3-0)^{2}+(2-5)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$ units

$C D=\sqrt{(0+3)^{2}+(-1-2)^{2}}=\sqrt{(3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$ units

$D A=\sqrt{(0-3)^{2}+(-1-2)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$ units

Therefore, $A B=B C=C D=D A=3 \sqrt{2}$ units

Also, $A C=\sqrt{(-3-3)^{2}+(2-2)^{2}}=\sqrt{(-6)^{2}+(0)^{2}}=\sqrt{36}=6$ units

$B D=\sqrt{(0-0)^{2}+(-1-5)^{2}}=\sqrt{(0)^{2}+(-6)^{2}}=\sqrt{36}=6$ units

Thus, diagonal $A C=$ diagonal $B D$

Therefore, the given points form a square.

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

$A B=\sqrt{(2-6)^{2}+(1-2)^{2}}=\sqrt{(-4)^{2}+(-1)^{2}}=\sqrt{16+1}=\sqrt{17}$ units

$B C=\sqrt{(1-2)^{2}+(5-1)^{2}}=\sqrt{(-1)^{2}+(-4)^{2}}=\sqrt{1+16}=\sqrt{17}$ units

$C D=\sqrt{(5-1)^{2}+(6-5)^{2}}=\sqrt{(4)^{2}+(1)^{2}}=\sqrt{16+1}=\sqrt{17}$ units

$D A=\sqrt{(5-6)^{2}+(6-2)^{2}}=\sqrt{(1)^{2}+(4)^{2}}=\sqrt{1+16}=\sqrt{17}$ units

Therefore $A B=B C=C D=D A=\sqrt{17}$ units

Also $A C=\sqrt{(1-6)^{2}+(5-2)^{2}}=\sqrt{(-5)^{2}+(3)^{2}}=\sqrt{25+9}=\sqrt{34}$ units

$B D=\sqrt{(5-2)^{2}+(6-1)^{2}}=\sqrt{(3)^{2}+(5)^{2}}=\sqrt{9+25}=\sqrt{34}$ units

Thus, diagonal $A C=$ diagonal $B D$

Therefore, the given points form a square.

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

$P Q=\sqrt{(3-0)^{2}+(1+2)^{2}}=\sqrt{(3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$ units

$Q R=\sqrt{(0-3)^{2}+(4-1)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$ units

$R S=\sqrt{(-3-0)^{2}+(1-4)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$ units

$S P=\sqrt{(-3-0)^{2}+(1+2)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$ units

Therefore, $P Q=Q R=R S=S P=3 \sqrt{2}$ units

Also, $P R=\sqrt{(0-0)^{2}+(4+2)^{2}}=\sqrt{(0)^{2}+(6)^{2}}=\sqrt{36}=6$ units

$Q S=\sqrt{(-3-3)^{2}+(1-1)^{2}}=\sqrt{(-6)^{2}+(0)^{2}}=\sqrt{36}=6$ units

Thus, diagonal $P R=$ diagonal $Q S$

Therefore, the given points form a square.