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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)


Solution

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

AB=(2(4))2+(4(1))2

=(2)2+(3)2

=4+9

=13

=13 units

BC=(4(2))2+(0(4))2

=(6)2+(4)2

=36+16

=52

=213 units

CD=(24)2+(30)2

=(2)2+(3)2

=4+9

=13

=13 units

AD=(2(4))2+(3(1))2

=(6)2+(4)2

=36+16

=52

=213 units

Thus AB=CD=13 units and BC=AD=213 units

Also,
AC=(4(4))2+(0(1))2

=(8)2+(1)2

=64+1

=65

=65 units

BD=(2(2))2+(3(4))2

=(4)2+(7)2

=16+49

=65

=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=(142)2+(10(2))2

=(12)2+(23)2

=144+144

=288

=122 units

BC=(1114)2+(1310)2

=(3)2+(3)2

=9+9

=18

=32 units

CD=(111)2+(113)2

=(12)2+(12)2

=144+144

=288

=122 units

AD=(12)2+(1(2))2

=(3)2+(3)2

=9+9

=18

=32 units

Thus AB=CD=122units and BC=AD=32 unit

Also,
AC=(112)2+(13(2))2

=(9)2+(15)2

=81+225

=306

=334 units

BD=(114)2+(110)2

=(15)2+(9)2

=81+225

=306

=334 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=(60)2+(2(4))2

=(6)2+(6)2

=36+36

=72

=62 units

BC=(36)2+(52)2

=(3)2+(3)2

=9+9

=18

=32 units

CD=(33)2+(15)2

=(6)2+(6)2

=36+36

=72

=62 units

AD=(30)2+(1(4))2

=(3)2+(3)2

=9+9

=18

=32 units

Thus AB=CD=62 units and BC=AD=32 units

Also,
AC=(30)2+(5(4))2

=(3)2+(9)2

=9+81

=90

=310 units

BD=(36)2+(12)2

=(9)2+(3)2

=81+9

=90

=310 units

Also, diagonal AC = diagonal BD

Hence, the given points form a rectangle.

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