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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 point(0,7,−10),(1,6,−6) and (4,9,−6) be denoted by A,B and C respectivelyAB=√(1−0)2+(6−7)2+(−6+10)2=√(1)2+(−1)2+(4)2=√1+1+16=√18⇒AB=3√2BC=√(4−1)2+(9−6)2+(−6+6)2=√(3)2+(3)2=√9+9=√18⇒BC=3√2CA=√(0−4)2+(7−9)2+(−10+6)2=√(−4)2+(−2)2+(−4)2=√16+4+16=√36=6Here AB=BC ≠ CAThus 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 respectivelyAB=√(−1−0)2+(6−7)2+(6−10)2=√(−1)2+(−1)2+(−4)2=√1+1+16=√18=3√2 BC=√(−4+1)2+(9−6)2+(6−6)2=√(−3)2+(3)2+(0)2=√9+9=√18=3√2CA=√(0+4)2+(7−9)2+(10−6)2=√(4)2+(−2)2+(4)2=√16+4+16=√36=6Now AB2+BC2=(3√2)2+(3√2)2=18+18=36=AC2Therefore by pythagoras theorem ABC is a right triangleHence 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 respectivelyAB=√(1+1)2+(−2−2)2+(5−1)2=√4+16+16AB=√36AB=6BC=√(4−1)2+(−7+2)2+(8−5)2=√9+25+9=√43CD=√(2−4)2+(−3+7)2+(4−8)2=√4+16+16=√36CD=6DA=√(−1−2)2+(2+3)2+(1−4)2DA=√9+25+9=√43Here AB=CD=6, BC=AD=√43Hence the opposite sides of quadrilateral ABCD whose vertices are taken in order are equalTherefore ABCD is a parallelogramHence the given points are the vertices of a parallelogram

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