S1 : If (0, 3), (1,1) and (–1, 2) be the mid points of the sides of a triangle, then centroid of the original triangle is (0, 2)
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Step-by-step explanation:
Let the vertices be (x
1
,y
1
),(x
2
,y
2
),(x
3
,y
3
)
Centroid of △=(
3
x
1
+x
2
+x
3
,
3
y
1
+y
2
+y
3
)
Midpoints of triangle are (
2
x
1
+x
2
,
2
y
1
+y
2
),(
2
x
2
+x
3
,
2
y
2
+y
3
)(
2
x
1
+x
3
,
2
y
1
+y
3
)
Hence centroid is
3
2
x
1
+x
2
+
2
x
2
+x
3
+
2
x
1
+x
3
,
3
2
y
1
+y
2
+
2
y
2
+y
3
+
2
y
1
+y
3
=
3
x
1
+x
2
+x
3
,
3
y
1
+y
2
+y
3
We observe that △ formed by midpoints also has same centroid.
∴ Centroid of original △=(
3
0+2+−2
,
3
3+2+1
)
=(0,2)
So both assertion and reason are explained
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