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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=√(1−0)2+(6−7)2+(−6+10)2 = √1+1+16=√18=3√2 BC=√(4−1)2+(9−6)2+(−6+6)2 = √9+9+0=√18=3√2 AC=√(4−0)2+(9−7)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=√(−1−0)2+(6−7)2+(6−10)2 = √1+1+16=3√2 BC=√(−4+1)2+(9−6)2+(6−6)2 = √9+9=0=√18=3√2 AC=√(−4−0)2+(9−7)2+(6−10)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+(−2−2)2+(5−1)2 = √4+16+16=√36=6 BC=√(4−1)2+(−7+2)2+(8−5)2 = √9+25+9=√43 CD=√(2−4)2+(−3+7)2+(4−8)2 = √4+16+16=√36=6 AD=√(2+1)2+(−3−2)2+(4−1)2 = √9+25+9=√43 AC=√(4+1)2+(−7−2)2+(8−1)2 = √25+81+49=√155 BD=√(2−1)2+(−3+2)2+(4−5)2 = √1+1+1=√3 Now AB=CD, BC=AD and AC≠BC Thus A, B, C and D are vertices of a parallelogram ABCD.

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