In a AXYZ, KA is the median, the vertices of the Axyz are
(5,2) (4,7) and (6,3) show that medias XA divides the given
triangle intog 2 equal areas
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Explanation:
Let AD is the median of △ABC. Then D is the midpoint of BC.
∴Co-ordinates of D=(
2
x
1
+x
2
,
2
y
1
+y
2
)=(
2
3+5
,
2
−2+2
)=(4,0)
Median AD divide the △ABC into two triangles.
∴Area of △ABD=
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
Here(x
1
,y
1
)=(4,−6)
(x
2
,y
2
)=(3,−2)
(x
3
,y
3
)=(4,0)
Area of △ABD=
2
1
[4(−2−0)+3(0+6)+4(−6+2)]
=
2
1
[−8+18−16]=
2
−6
=3
∴ Area of △ADC=
2
1
[x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)]
Here(x
1
,y
1
)=(4,−6)
(x
2
,y
2
)=(4,0)
(x
3
,y
3
)=(5,−2)
Area of △ADC=
2
1
[4(0+2)+4(−2+6)+5(−6−0)]
=
2
1
[−8+16−30]=
2
−6
=3
∴ Area of △ABD=Area of △ADC
Hence it is proved that a median of a triangle divides it into two triangles of equal areas.
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