Question

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

Answer 1

(i) Let point$(0,7,-10),(1,6,-6)$ and $(4,9,-6)$ be denoted by $A,B$ and $C$ respectively
$AB=\sqrt{{(1-0)}^2+{(6-7)}^2+{(-6+10)}^2}$
$=\sqrt{{\left(1\right)}^2+{(-1)}^2+{\left(4\right)}^2}$
$=\sqrt{1+1+16}$
$=\sqrt{18}$
$\Rightarrow AB=3\sqrt{2}$
$BC=\sqrt{{(4-1)}^2+{(9-6)}^2+{(-6+6)}^2}$
=$\sqrt{{\left(3\right)}^2+{\left(3\right)}^2}$
=$\sqrt{9+9}=\sqrt{18}$
$\Rightarrow BC=3\sqrt{2}$
$CA=\sqrt{{(0-4)}^2+{(7-9)}^2+{(-10+6)}^2}$
=$\sqrt{{(-4)}^2+{(-2)}^2+{(-4)}^2}$
=$\sqrt{16+4+16}=\sqrt{36}=6$
Here $AB=BC$ $\ne$ $CA$
Thus the given points are the vertices of an isosceles triangle
(ii) Let $(0,7,10),(-1,6,6)$ and $(-4,9,6)$ be denoted by $A,B$ and $C$ respectively
$AB=\sqrt{{(-1-0)}^2+{(6-7)}^2+{(6-10)}^2}$
$=\sqrt{{(-1)}^2+{(-1)}^2+{(-4)}^2}$
$=\sqrt{1+1+16}=\sqrt{18}$
$=3\sqrt{2}$
$BC=\sqrt{{(-4+1)}^2+{(9-6)}^2+{(6-6)}^2}$
$=\sqrt{{(-3)}^2+{\left(3\right)}^2+{\left(0\right)}^2}$
$=\sqrt{9+9}=\sqrt{18}$
$=3\sqrt{2}$
$CA=\sqrt{{(0+4)}^2+{(7-9)}^2+{(10-6)}^2}$
$=\sqrt{{\left(4\right)}^2+{(-2)}^2+{\left(4\right)}^2}$
$=\sqrt{16+4+16}$
$=\sqrt{36}$
$=6$
Now $A{B}^2+B{C}^2={\left(3\sqrt{2}\right)}^2+{\left(3\sqrt{2}\right)}^2=18+18=36=A{C}^2$
Therefore by pythagoras theorem $ABC$ is a right triangle
Hence the given points are the vertices of a right-angled triangle
(iii) Let $(-1,2,1),(1,-2,5),(4,-7,8)$ and $(2,-3,4)$ be denoted by $A,B,C$ and $D$ respectively
$AB=\sqrt{{(1+1)}^2+{(-2-2)}^2+{(5-1)}^2}$
$=\sqrt{4+16+16}$
$AB=\sqrt{36}$
$AB=6$
$BC=\sqrt{{(4-1)}^2+{(-7+2)}^2+{(8-5)}^2}$
$=\sqrt{9+25+9}=\sqrt{43}$
$CD=\sqrt{{(2-4)}^2+{(-3+7)}^2+{(4-8)}^2}$
$=\sqrt{4+16+16}$
$=\sqrt{36}$
$CD=6$
$DA=\sqrt{{(-1-2)}^2+{(2+3)}^2+{(1-4)}^2}$
$DA=\sqrt{9+25+9}=\sqrt{43}$
Here $AB=CD=6$, $BC=AD=\sqrt{43}$
Hence the opposite sides of quadrilateral $ABCD$ whose vertices are taken in order are equal
Therefore $ABCD$ is a parallelogram
Hence the given points are the vertices of a parallelogram