The co-ordinates of the mid-points of the sides BC, CA and AB of triangles ABC are
(3, 4), 2, 21 and (5, 0) ( ) resepectively. If D is the mid-point of BC then the ∆ABD is
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Answered by
1
Answer:
Let A=(x
1
,y
1
,z
1
),B=(x
2
,y
2
,z
2
),C=(x
3
,y
3
,z
3
)
From the figure,
x
1
+x
2
=2l,y
1
+y
2
=0,z
1
+z
2
=0, [midpoint formula]
x
2
+x
3
=0,y
2
+y
3
=2m,z
2
+z
3
=0
and x
1
+x
3
=0,y
1
+y
3
=0,z
1
+z
3
=2n
On solving, we get
x
1
=l,x
2
=l,x
3
=−l,
y
1
=−m,y
2
=m,y
3
=m
and z
1
=n,z
2
=−n,z
3
=n
∴ Coordinates are A(l,−m,n),B(l,m,−n) and C(−l,m,n)
∴
l
2
+m
2
+n
2
AB
2
+BC
2
+CA
2
=
l
2
+m
2
+n
2
4m
2
+4n
2
+4l
2
+4n
2
+(4l
2
+4m
2
)
=8
Answered by
1
Step-by-step explanation:
sorry in not under stand about this sorry
sorry sorry don't mis understand
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