are (-3,1),(2,6) & (7,-3) collinear
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Answer:
We have,
Points are
A(x
1
,y
1
,z
1
)=(1,2,7)
B(x
2
,y
2
,z
2
)=(2,6,3)
C(x
3
,y
3
,z
3
)=(3,10,−1)
Now,
Using distance formula
AB=
(x
1
−x
2
)
2
+(y
1
−y
2
)
2
+(z
1
−z
2
)
2
AB=
(1−2)
2
+(2−6)
2
+(7−3)
2
AB=
1+16+16
AB=
33
BC=
(x
2
−x
3
)
2
+(y
2
−y
3
)
2
+(z
2
−z
3
)
2
BC=
(2−3)
2
+(6−10)
2
+(3+1)
2
BC=
1+16+16
BC=
33
CA=
(x
3
−x
1
)
2
+(y
3
−y
1
)
2
+(z
3
−z
1
)
2
CA=
(3−1)
2
+(10−2)
2
+(−1−7)
2
CA=
4+64+64
CA=
132
CA=
2×2×33
CA=2
33
Then,
CA=AB+BC
2
33
=
33
+
33
2
33
=2
33
Then,
It's points are collinear.
Hence, this is the answer.
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