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

Solution:

(i)

Given : A $(-4,-1)$, B $(-2,-4)$, C $(4,0)$ and D $(2,3)$

In rectangle the opposite sides are equal

Using $d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

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

$\mathrm{AB}=\sqrt{13}$

$\mathrm{BC}=\sqrt{(4+2)^{2}+(0+4)^{2}}$

$\mathrm{BC}=\sqrt{52}$

$\mathrm{CD}=\sqrt{(2-4)^{2}+(3-0)^{2}}$

$\mathrm{CD}=\sqrt{13}$

$\mathrm{AD}=\sqrt{(2+4)^{2}+(3+1)^{2}}$

$\mathrm{AD}=\sqrt{52}$

$\mathrm{AB}=\mathrm{CD}$ and $\mathrm{BC}=\mathrm{AD}$

Now, we will prove for rectangles because diagnols in a rectangle are also equal

$\mathrm{AC}=\sqrt{(4-(-4))^{2}+(0-(-1))^{2}}$

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

$\mathrm{AC}=\sqrt{65}$

And

$\mathrm{BD}=\sqrt{(2-(-2))^{2}+(3-(-4))^{2}}$

$\mathrm{BD}=\sqrt{(4)^{2}+(7)^{2}}$

$\mathrm{BD}=\sqrt{65}$

We can see that diagonal $s$ are also equal

Therefore, given coordinates are of rectangle.

(ii) The given points are A(2, −2), B(14, 10) C(11, 13) and D(−1, 1).

$A B=\sqrt{(14-2)^{2}+\{10-(-2)\}^{2}}=\sqrt{(12)^{2}+(12)^{2}}=\sqrt{144+144}=\sqrt{288}=12 \sqrt{2}$ units

$B C=\sqrt{(11-14)^{2}+(13-10)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$ units

$C D=\sqrt{(-1-11)^{2}+(1-13)^{2}}=\sqrt{(-12)^{2}+(-12)^{2}}=\sqrt{144+144}=\sqrt{288}=12 \sqrt{2}$ units

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

Thus, $A B=C D=12 \sqrt{2}$ units and $B C=A D=3 \sqrt{2}$ units

Also, $A C=\sqrt{(11-2)^{2}+\{13-(-2)\}^{2}}=\sqrt{(9)^{2}+(15)^{2}}=\sqrt{81+225}=\sqrt{306}=3 \sqrt{34}$ units

$B D=\sqrt{(-1-14)^{2}+(1-10)^{2}}=\sqrt{(-15)^{2}+(-9)^{2}}=\sqrt{81+225}=\sqrt{306}=3 \sqrt{34}$ units

Also, diagonal AC = diagonal BD
Hence, the given points form a rectangle.

(iii) The given points are A(0, −4), B(6, 2) C(3, 5) and D(−3, −1).

$A B=\sqrt{(6-0)^{2}+\{2-(-4)\}^{2}}=\sqrt{(6)^{2}+(6)^{2}}=\sqrt{36+36}=\sqrt{72}=6 \sqrt{2}$ units

$B C=\sqrt{(3-6)^{2}+(5-2)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$ units

$C D=\sqrt{(-3-3)^{2}+(-1-5)^{2}}=\sqrt{(-6)^{2}+(-6)^{2}}=\sqrt{36+36}=\sqrt{72}=6 \sqrt{2}$ units

 

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

Thus, $A B=C D=\sqrt{10}$ units and $B C=A D=\sqrt{5}$ units

Also, $A C=\sqrt{(3-0)^{2}+\{5-(-4)\}^{2}}=\sqrt{(3)^{2}+(9)^{2}}=\sqrt{9+81}=\sqrt{90}=3 \sqrt{10}$ units

$B D=\sqrt{(-3-6)^{2}+(-1-2)^{2}}=\sqrt{(-9)^{2}+(-3)^{2}}=\sqrt{81+9}=\sqrt{90}=3 \sqrt{10}$ units

Also, diagonal AC = diagonal BD.
Hence, the given points form a rectangle.