Verify the following: (0,7,-10), (1,6,-6) and (4,9,-6) are vertices of an isosceles triangle.

Answer

N/a

Explanation

(i) Let A=(0,7,-10),B=(1,6,-6) and C=(4,9,-6)
`therefore AB=sqrt((1-0)^(2)+(6-7)^(2)+(-6+10)^(2))`
`=sqrt(1+1+16)=sqrt(18)=3sqrt(2)`
`BC=sqrt((4-1)^(2)+(9-6)^(2)+(-6+6)^(2))`
`=sqrt(9+9+0)=sqrt(18)+3sqrt(2)`
and `CA=sqrt((0-4)^(2)+(7-9)^(2)+(-10+6)^(2))`
`=sqrt(16+4+16)=sqrt(36)=6`
`because AB=BC`
`therefore DeltaABC` is isosceles triangle.
(ii) Let `A-=(0,7,10),B-=(-1,6,6) " and" C-=(-4,9,6)`
`therefore AB^(2)=(-1-0)^(2)+(6-7)^(2)+(6-10)^(2)=1+1+16=18`
`BC^(2)=(-4+1)^(2)+(9-6)^(2)+(6-6)^(2)=9+9+0=18`
and `CA^(2)=(0+4)^(2)+(7-9)^(2)+(10-6)^(2)`
=16+4+16=36
Now, `AB^(2)+BC^(2)=18+18=36=CA^(2)`
`implies DeltaABC` is right -angled triangle.
(iii) Let `A-=(-1,2,1),B-=(1,-2,5),C-=(4,-7,8) " and" D-=(2,-3,4)`
`therefore 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`
and `DA=sqrt((-1-2)^(2)+(2+3)^(2)+(1-4)^(2))`
`=sqrt(9+25+9)=sqrt(43)`
`therefore AB=CD ` and BC=DA
`therefore ABCD` is a parallelogram.