Question

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

(i) A(-4, -1), B (-2, -4), C(4, 0) and D(2, 3)
(ii) A(2, -2), B(14, 10), C(11, 13) and D(-1, 1)
(iii) A(0, -4), B(6, 2), C(3, 5) and D(-3, -1)

Answer 1

(i) The given points are A (-4,-1), B(-2,-4) and C(4,0), D(2,3). Then

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

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

$=\sqrt{4+9}$

$=\sqrt{13}$

$=\sqrt{13}$ units

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

$=\sqrt{(6{)}^2+(4{)}^2}$

$=\sqrt{36+16}$

$=\sqrt{52}$

$=2\sqrt{13}$ units

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

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

$=\sqrt{4+9}$

$=\sqrt{13}$

$=\sqrt{13}$ units

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

$=\sqrt{(6{)}^2+(4{)}^2}$

$=\sqrt{36+16}$

$=\sqrt{52}$

$=2\sqrt{13}$ units

Thus $AB=CD=\sqrt{13}$ units and $BC=AD=2\sqrt{13}$ units

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

$=\sqrt{(8{)}^2+(1{)}^2}$

$=\sqrt{64+1}$

$=\sqrt{65}$

$=\sqrt{65}$ units

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

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

$=\sqrt{16+49}$

$=\sqrt{65}$

$=\sqrt{65}$ units

Also, diagonal AC = diagonal BD

Hence, the given points form a rectangle.

(ii) A (2,-2), B(14,10) and C(11,13), D(-1,1).

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

$AB=\sqrt{(14-2{)}^2+(10-(-2){)}^2}$

$=\sqrt{(12{)}^2+(23{)}^2}$

$=\sqrt{144+144}$

$=\sqrt{288}$

$=12\sqrt{2}$ units

$BC=\sqrt{(11-14{)}^2+(13-10{)}^2}$

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

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$ units

$CD=\sqrt{(-1-11{)}^2+(1-13{)}^2}$

$=\sqrt{(-12{)}^2+(-12{)}^2}$

$=\sqrt{144+144}$

$=\sqrt{288}$

$=12\sqrt{2}$ units

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

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

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$ units

Thus $AB=CD=12\sqrt{2}$units and $BC=AD=3\sqrt{2}$ unit

Also,
$AC=\sqrt{(11-2{)}^2+(13-(-2){)}^2}$

$=\sqrt{(9{)}^2+(15{)}^2}$

$=\sqrt{81+225}$

$=\sqrt{306}$

$=3\sqrt{34}$ units

$BD=\sqrt{(-1-14{)}^2+(1-10{)}^2}$

$=\sqrt{(-15{)}^2+(-9{)}^2}$

$=\sqrt{81+225}$

$=\sqrt{306}$

$=3\sqrt{34}$ units

Also, diagonal AC = diagonal BD

Hence, the given points form a rectangle.

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

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

$AB=\sqrt{(6-0{)}^2+(2-(-4){)}^2}$

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

$=\sqrt{36+36}$

$=\sqrt{72}$

$=6\sqrt{2}$ units

$BC=\sqrt{(3-6{)}^2+(5-2{)}^2}$

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

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$ units

$CD=\sqrt{(-3-3{)}^2+(-1-5{)}^2}$

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

$=\sqrt{36+36}$

$=\sqrt{72}$

$=6\sqrt{2}$ units

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

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

$=\sqrt{9+9}$

$=\sqrt{18}$

$=3\sqrt{2}$ units

Thus $AB=CD=6\sqrt{2}$ units and $BC=AD=3\sqrt{2}$ units

Also,
$AC=\sqrt{(3-0{)}^2+(5-(-4){)}^2}$

$=\sqrt{(3{)}^2+(9{)}^2}$

$=\sqrt{9+81}$

$=\sqrt{90}$

$=3\sqrt{10}$ units

$BD=\sqrt{(-3-6{)}^2+(-1-2{)}^2}$

$=\sqrt{(-9{)}^2+(-3{)}^2}$

$=\sqrt{81+9}$

$=\sqrt{90}$

$=3\sqrt{10}$ units

Also, diagonal AC = diagonal BD

Hence, the given points form a rectangle.