Question

Verify the following:
(i) (0, 7, –10), (1, 6, –6) and (4, 9, –6) are vertices of an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6) and (–4, 9, –6) are vertices of a right-angled triangle.
(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram.

Answer 1

(i) Let A(0, 7, $-$10) , B(1, 6, $-$6) , C(4, 9, $-$6) be the vertices of $△ABC$.Then,
AB = $\sqrt{{\left(1-0\right)}^2+{\left(6-7\right)}^2+{\left(-6+10\right)}^2}$
$$\begin{gathered} =\sqrt{1^2+{\left(-1\right)}^2+4^2} \\ =\sqrt{1+1+16} \\ =\sqrt{18} \\ =3\sqrt{2} \end{gathered}$$

BC = $\sqrt{{\left(4-1\right)}^2+{\left(9-6\right)}^2+{\left(-6+6\right)}^2}$
$$\begin{gathered} =\sqrt{3^2+3^2+0} \\ =\sqrt{9+9} \\ =\sqrt{18} \\ =3\sqrt{2} \end{gathered}$$

CA= $\sqrt{{\left(0-4\right)}^2+{\left(7-9\right)}^2+{\left(-10+6\right)}^2}$
$$\begin{gathered} =\sqrt{16+4+16} \\ =\sqrt{36} \\ =6 \end{gathered}$$
Clearly, AB = BC
Thus, the given points are the vertices of an isosceles triangle.


(ii) Let A(0,7,10) , B( $-$1,6,6) and C( $-$4,9,6) be the vertices of $△ABC$. Then ,

AB = $\sqrt{{\left(-1-0\right)}^2+{\left(6-7\right)}^2+{\left(6-10\right)}^2}$

$$\begin{gathered} =\sqrt{{\left(-1\right)}^2+{\left(-1\right)}^2+{\left(-4\right)}^2} \\ =\sqrt{1+1+16} \\ =\sqrt{18} \\ =3\sqrt{2} \end{gathered}$$

BC = $\sqrt{{\left(-4+1\right)}^2+{\left(9-6\right)}^2+{\left(6-6\right)}^2}$
$$\begin{gathered} =\sqrt{{\left(-3\right)}^2+{\left(3\right)}^2+0} \\ =\sqrt{9+9} \\ =\sqrt{18} \\ =3\sqrt{2} \end{gathered}$$

AC = $\sqrt{{\left(-4-0\right)}^2+{\left(9-7\right)}^2+{\left(6-10\right)}^2}$
$$\begin{gathered} =\sqrt{{\left(-4\right)}^2+{\left(2\right)}^2+{\left(-4\right)}^2} \\ =\sqrt{16+4+16} \\ =\sqrt{36} \\ =6 \end{gathered}$$
$A{C}^2$$=A{B}^2+B{C}^2$
Thus, the given points are the vertices of a right-angled triangle.

(iii) Let A($-$1, 2, 1) , B(1, $-$2, 5) , C(4, $-$7, 8), D(2, $-$3, 4) be the vertices of quadrilateral $\square ABCD$

$$\begin{gathered} AB=\sqrt{{\left(1+1\right)}^2+{\left(-2-2\right)}^2+{\left(5-1\right)}^2} \\ =\sqrt{4+16+16} \\ =\sqrt{36} \\ =6 \\ BC=\sqrt{{\left(4-1\right)}^2+{\left(-7+2\right)}^2+{\left(8-5\right)}^2} \\ =\sqrt{9+25+9} \\ =\sqrt{43} \\ CD=\sqrt{{\left(2-4\right)}^2+{\left(-3+7\right)}^2+{\left(4-8\right)}^2} \\ =\sqrt{4+16+16} \\ =\sqrt{36} \\ =6 \\ DA=\sqrt{{\left(-1-2\right)}^2+{\left(2+3\right)}^2+{\left(1-4\right)}^2} \\ =\sqrt{9+25+9} \\ =\sqrt{43} \\ \therefore AB=CD\mathrm{and}BC=DA \end{gathered}$$
Since, each pair of opposite sides are equal.
Thus, quadrilateral $\square ABCD$ is a parallelogram.