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 $A(0,7,-10),B(1,6,-6)$ and $C(4,9,-6)$ be three vertices of triangle ABC. Then

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

= $\sqrt{1+1+16}=\sqrt{18}=3\sqrt{2}$

$BC=\sqrt{(4-1{)}^2+(9-6{)}^2+(-6+6{)}^2}$

= $\sqrt{9+9+0}=\sqrt{18}=3\sqrt{2}$

$AC=\sqrt{(4-0{)}^2+(9-7{)}^2+(-6+10{)}^2}$

= $\sqrt{16+4+16}=\sqrt{36}=6$

Now $AB=BC$

Thus, ABC is an isosceles triangle.

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

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

= $\sqrt{1+1+16}=3\sqrt{2}$

$BC=\sqrt{(-4+1{)}^2+(9-6{)}^2+(6-6{)}^2}$

= $\sqrt{9+9=0}=\sqrt{18}=3\sqrt{2}$

$AC=\sqrt{(-4-0{)}^2+(9-7{)}^2+(6-10{)}^2}$

= $\sqrt{16+4+16}=\sqrt{36}=6$

Now $A{C}^2=A{B}^2+B{C}^2$

Thus, ABC is a right angled triangle

(iii) Let $A(-1,2,1),B(1,-2,5)$ and $C(4,-7,8)$ and $D(2,-3,4)$ be four vertices of a quadrilateral ABCD. Then

$AB=\sqrt{(1+1{)}^2+(-2-2{)}^2+(5-1{)}^2}$

= $\sqrt{4+16+16}=\sqrt{36}=6$

$BC=\sqrt{(4-1{)}^2+(-7+2{)}^2+(8-5{)}^2}$

= $\sqrt{9+25+9}=\sqrt{43}$

$CD=\sqrt{(2-4{)}^2+(-3+7{)}^2+(4-8{)}^2}$

= $\sqrt{4+16+16}=\sqrt{36}=6$

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

= $\sqrt{9+25+9}=\sqrt{43}$

$AC=\sqrt{(4+1{)}^2+(-7-2{)}^2+(8-1{)}^2}$

= $\sqrt{25+81+49}=\sqrt{155}$

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

= $\sqrt{1+1+1}=\sqrt{3}$

Now $AB=CD,BC=AD$ and $AC\ne BC$

Thus A, B, C and D are vertices of a parallelogram ABCD.