let a(-a,0), b(0,a) and c (a,beta) be the vertices of the triangle abc and g be its centroid. prove that ga²+gb²+gc²=1/3 (ab²+bc²+ca²)
Answers
Answer:
LetA(x1,y1),B(x2,y2)andC(x3,y3)betheverticesof△ABC.
Withoutthelawofgenerality,assumethecentroidofthe
△ABCtobeatorigin,i.e.,G=(0,0).
Centroidof△ABC=[
3
x1+x2+x3
,
3
y1+y2+y3
]
∴x1+x2+x3=0;y1+y2+y3=0
Squaringonbothsides,
x1
2
+x2
2
+x3
2
+2x1.x2+2x2.x3+2x3.x1=0
y1
2
+y2
2
+y3
2
+2y1.y2+2y2.y3+2y3.y1=0−(i)
AB
2
+BC
2
+CA
2
=[(x2−x1)
2
+(y2−y1)
2
]+[(x3−x2)
2
+(y3−y2)
2
]+[(x1−x3)
2
+(y1−y3)
2
]
=(x1
2
+x2
2
−2x1x2+y1
2
+y2
2
−2y1y2)+(x1
2
+x3
2
−2x1x3+y1
2
+y3
2
−2y2y3)
+(x1
2
+x3
2
−2x1x3+y1
2
+y3
2
−2y1y3)
=(2x1
2
+2x2
2
+2x3
2
−2x1x2−2x2x3−2x1x3)+
(2y1
2
+2y2
2
+2y3
2
−y1y2−2y2y3−2y1y3)
=(3x1
2
+3x2
2
+3x3
2
)+(3y1
2
+3y2
2
+3y3
2
)
=3(x1
2
+x2
2
+x3
2
)+3(y1
2
+y2
2
+y3
2
)−(ii)
3(GA
2
+GB
2
+GC
2
)
=3[(x1−0)
2
+(y1−0)
2
+(x2−0)
2
+(y2−0)
2
+(x3−0)
2
+(y3−0)
2
]
=3[x1
2
+y1
2
+x2
2
+y2
2
+x3
2
+y3
2
]
=3(x1
2
+x2
2
+x3
2
)+3(y1
2
+y2
2
+y3
2
)−(iii)
from(ii)&(iii)
AB
2
+BC
2
+CA
2
=3(GA
2
+GB
2
+GC
2
)