In the given example if 0 = 7 thenΔΟ ΔΟ+ ΟΔΟΟΔΕ9 1 3 2 7 2what is the value of A, serially?1) 3,52) 5,33) 4,64) 6.4 About the author Piper
Answer: (i) Let point(0,7,−10),(1,6,−6) and (4,9,−6) be denoted by A,B and C respectively AB= (1−0) 2 +(6−7) 2 +(−6+10) 2 = (1) 2 +(−1) 2 +(4) 2 = 1+1+16 = 18 ⇒AB=3 2 BC= (4−1) 2 +(9−6) 2 +(−6+6) 2 = (3) 2 +(3) 2 = 9+9 = 18 ⇒BC=3 2 CA= (0−4) 2 +(7−9) 2 +(−10+6) 2 = (−4) 2 +(−2) 2 +(−4) 2 = 16+4+16 = 36 =6 Here AB=BC = CA Thus 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 respectively AB= (−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 2 CA= (0+4) 2 +(7−9) 2 +(10−6) 2 = (4) 2 +(−2) 2 +(4) 2 = 16+4+16 = 36 =6 Now AB 2 +BC 2 =(3 2 ) 2 +(3 2 ) 2 =18+18=36=AC 2 Therefore by pythagoras theorem ABC is a right triangle Hence 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 respectively AB= (1+1) 2 +(−2−2) 2 +(5−1) 2 = 4+16+16 AB= 36 AB=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 CD=6 DA= (−1−2) 2 +(2+3) 2 +(1−4) 2 DA= 9+25+9 = 43 Here AB=CD=6, BC=AD= 43 Hence the opposite sides of quadrilateral ABCD whose vertices are taken in order are equal Therefore ABCD is a parallelogram Hence the given points are the vertices of a parallelogram Reply
Answer:
(i) Let point(0,7,−10),(1,6,−6) and (4,9,−6) be denoted by A,B and C respectively
AB=
(1−0)
2
+(6−7)
2
+(−6+10)
2
=
(1)
2
+(−1)
2
+(4)
2
=
1+1+16
=
18
⇒AB=3
2
BC=
(4−1)
2
+(9−6)
2
+(−6+6)
2
=
(3)
2
+(3)
2
=
9+9
=
18
⇒BC=3
2
CA=
(0−4)
2
+(7−9)
2
+(−10+6)
2
=
(−4)
2
+(−2)
2
+(−4)
2
=
16+4+16
=
36
=6
Here AB=BC
= CA
Thus 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 respectively
AB=
(−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
2
CA=
(0+4)
2
+(7−9)
2
+(10−6)
2
=
(4)
2
+(−2)
2
+(4)
2
=
16+4+16
=
36
=6
Now AB
2
+BC
2
=(3
2
)
2
+(3
2
)
2
=18+18=36=AC
2
Therefore by pythagoras theorem ABC is a right triangle
Hence 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 respectively
AB=
(1+1)
2
+(−2−2)
2
+(5−1)
2
=
4+16+16
AB=
36
AB=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
CD=6
DA=
(−1−2)
2
+(2+3)
2
+(1−4)
2
DA=
9+25+9
=
43
Here AB=CD=6, BC=AD=
43
Hence the opposite sides of quadrilateral ABCD whose vertices are taken in order are equal
Therefore ABCD is a parallelogram
Hence the given points are the vertices of a parallelogram