Find the Centroid of the triangle whose vertices are x(1,4),y(-1,1)&z(3,-2).
Answers
\large{\red{\bold{\underline{Given:}}}}
Given:
\sf \: Coordinates \: of \: the \: vertices \: are: \: x(1,4), \: y(-1,1) \: and \: z(3,-2)Coordinatesoftheverticesare:x(1,4),y(−1,1)andz(3,−2)
\large{\green{\bold{\underline{To \: Find:}}}}
ToFind:
\sf \: Centroid \: of \: the \: triangleCentroidofthetriangle
\begin{gathered}\large{\blue{\bold{\underline{Formula \: Used:}}}} \\ \\ \sf \:Coordinates \: of \:Centroid = \: (\frac{x_{1} + x_{2} + x_{3}}{3} ),( \frac{y_{1} + y_{2} + y_{3}}{3})\end{gathered}
FormulaUsed:
CoordinatesofCentroid=(
3
x
1
+x
2
+x
3
),(
3
y
1
+y
2
+y
3
)
\begin{gathered}\large{\red{\bold{\underline{Solution:}}}} \\ \\ \: \sf \: On \: considering \: the \: respective \: coordinates \: as :\end{gathered}
Solution:
Onconsideringtherespectivecoordinatesas:
\begin{gathered}\sf \: x(1,4) \: arrow \: (x_{1}, y_{1}) \\ \\\sf \: y(-1,1) \: arrow \: (x_{2}, y_{2}) \\ \\ \sf \: z(3,-2) \: arrow \: (x_{3}, y_{3})\end{gathered}
x(1,4)arrow(x
1
,y
1
)
y(−1,1)arrow(x
2
,y
2
)
z(3,−2)arrow(x
3
,y
3
)
\begin{gathered}\large{\pink{\bold{\underline{Now:}}}} \\ \\ arrow \: \sf \: Centroid = ( \frac{1 + ( - 1) + 3}{3}) ,( \frac{4 + 1 - 2}{3} ) \\ \\ arrow \sf \: Centroid = ( \frac{0 + 3}{3}) , (\frac{5 - 2}{3} ) \\ \\ arrow \sf \: Centroid = ( \frac{3}{3} ),( \frac{3}{3} ) \\ \\ arrow \sf \: Centroid = (1,1)\end{gathered}
Now:
arrowCentroid=(
3
1+(−1)+3
),(
3
4+1−2
)
arrowCentroid=(
3
0+3
),(
3
5−2
)
arrowCentroid=(
3
3
),(
3
3
)
arrowCentroid=(1,1)
\begin{gathered}\large{\orange{\bold{\underline{Therefore:}}}} \\ \\ \sf \: The \: coordinates \: of \: Centroid \: of \: the \: triangle \\ \sf \: is \: (1,1).\end{gathered}
Therefore:
ThecoordinatesofCentroidofthetriangle
is(1,1).