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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.

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Solution

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

AB=(10)2+(67)2+(6+10)2

= 1+1+16=18=32

BC=(41)2+(96)2+(6+6)2

= 9+9+0=18=32

AC=(40)2+(97)2+(6+10)2

= 16+4+16=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=(10)2+(67)2+(610)2

= 1+1+16=32

BC=(4+1)2+(96)2+(66)2

= 9+9=0=18=32

AC=(40)2+(97)2+(610)2

= 16+4+16=36=6

Now AC2=AB2+BC2

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=(1+1)2+(22)2+(51)2

= 4+16+16=36=6

BC=(41)2+(7+2)2+(85)2

= 9+25+9=43

CD=(24)2+(3+7)2+(48)2

= 4+16+16=36=6

AD=(2+1)2+(32)2+(41)2

= 9+25+9=43

AC=(4+1)2+(72)2+(81)2

= 25+81+49=155

BD=(21)2+(3+2)2+(45)2

= 1+1+1=3

Now AB=CD, BC=AD and ACBC

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


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