Question

Show that the points A(1, 3, 4), B)-1, 6, 10), C(-7, 4, 7) and D(-5, 1, 1) are the vertices of a rhombus.

Answer 1

Let A(1, 3, 4), B)-1, 6, 10), C(-7, 4, 7) and D(-5, 1, 1) be the vertices of quadrilateral ABCD

AB=$\sqrt{(-1-1{)}^2+(6-3{)}^2+(10-4{)}^2}$

=$\sqrt{4+9+36}=\sqrt{49}$

=7 units

BC=$\sqrt{(-7+1{)}^2+(4-6{)}^2+(7-10{)}^2}$

=$\sqrt{36+4+9}=\sqrt{49}$

=7 units

CD=$\sqrt{(-5+7{)}^2+(1-4{)}^2+(1-7{)}^2}$

=$\sqrt{4+9+36}=\sqrt{49}$

=7 units

DA=$\sqrt{(1+5{)}^2+(3-1{)}^2+(4-1{)}^2}$

=$\sqrt{36+4+9}=\sqrt{49}$

=7 units $\therefore$

AB = BC = CD = DA

Hence, ABCD is a rhombus.