verify
(-14)×[(-3)+(-7)] = [(-14)×(-3)]+[-14×(-7)]
Answers
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
Answer:
(-14)×[(-3)+(-7)]= [(-14)×(-3)]+[-14×(-7)]=
=(-14)×[(-3-7)]=[(42)]+[98]
=(-14)×[-10]=[140]
= [140]=[140]
= 140=140
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