Question

The name of the figure formed by the points $(3,-5,1),(-1,0,8)$ and $(7,-10,-6)$ is

• A. a triangle
• B. a straight line
• C. an isosceles triangle
• D. an equilateral triangle

Answer 1

Answer: (B)

a straight line

Let $A=(3,-5,1),B=(-1,0,8)$ and $C=(7,-10,-6)$

Now,

$AB=\sqrt{(3+1{)}^2+(-5-0{)}^2+(1-8{)}^2}=\sqrt{90}=3\sqrt{10}$,

$BC=\sqrt{(-1-7{)}^2+(0+10{)}^2+(8+6{)}^2}=\sqrt{360}=6\sqrt{10}$

and $CA=\sqrt{(7-3{)}^2+(-10+5{)}^2+(-6-1{)}^2}=\sqrt{90}=3\sqrt{10}$

Clearly $BC=AB+CA$

$\therefore$ given points lies on straight line.