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 =2√13 units CD=√(2−4)2+(3−0)2 =√(−2)2+(3)2 =√4+9 =√13 =√13 units AD=√(2−(−4))2+(3−(−1))2 =√(6)2+(4)2 =√36+16 =√52 =2√13 units Thus AB=CD=√13 units and BC=AD=2√13 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=√(14−2)2+(10−(−2))2 =√(12)2+(23)2 =√144+144 =√288 =12√2 units BC=√(11−14)2+(13−10)2 =√(−3)2+(3)2 =√9+9 =√18 =3√2 units CD=√(−1−11)2+(1−13)2 =√(−12)2+(−12)2 =√144+144 =√288 =12√2 units AD=√(−1−2)2+(1−(−2))2 =√(−3)2+(3)2 =√9+9 =√18 =3√2 units Thus AB=CD=12√2units and BC=AD=3√2 unit Also, AC=√(11−2)2+(13−(−2))2 =√(9)2+(15)2 =√81+225 =√306 =3√34 units BD=√(−1−14)2+(1−10)2 =√(−15)2+(−9)2 =√81+225 =√306 =3√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=√(6−0)2+(2−(−4))2 =√(6)2+(6)2 =√36+36 =√72 =6√2 units BC=√(3−6)2+(5−2)2 =√(−3)2+(3)2 =√9+9 =√18 =3√2 units CD=√(−3−3)2+(−1−5)2 =√(−6)2+(−6)2 =√36+36 =√72 =6√2 units AD=√(−3−0)2+(−1−(−4))2 =√(−3)2+(3)2 =√9+9 =√18 =3√2 units Thus AB=CD=6√2 units and BC=AD=3√2 units Also, AC=√(3−0)2+(5−(−4))2 =√(3)2+(9)2 =√9+81 =√90 =3√10 units BD=√(−3−6)2+(−1−2)2 =√(−9)2+(−3)2 =√81+9 =√90 =3√10 units Also, diagonal AC = diagonal BD Hence, the given points form a rectangle.

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