The co-ordinates of centroid of a triangle having vertices $P (x_{1}, y_{1}), Q (x_{2}, y_{2})$ and $R (x_{3}, y_{3})$ are

(x_{1} + x_{2} + x_{3}, y_{1} + y_{2} + y_{2})

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(\begin{matrix} \frac{x_{1} + y_{1} + y_{2}}{3} \frac{x_{2} + y_{3} + x_{3}}{3} \end{matrix})

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(\begin{matrix} \frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{1} + y_{3}}{3} \end{matrix})

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None of the above

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Solution

The correct option is C $(\begin{matrix} \frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{1} + y_{3}}{3} \end{matrix})$
$x = \frac{x_{1} + x_{2} + x_{3}}{3}, y = \frac{y_{1} + y_{2} + y_{3}}{3}$
So, option C is correct.

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Similar questions

Find the centroid of the triangle with vertices A $(x_{1}, y_{1})$ , B $(x_{2}, y_{2})$ and C $(x_{3}, y_{3})$ .

Q. If three points

(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})

lie on the same line, prove that

\frac{y_{2} - y_{3}}{x_{2} x_{3}} + \frac{y_{3} - y_{1}}{x_{3} x_{1}} + \frac{y_{1} - y_{2}}{x_{1} x_{2}} = 0

Q. If

x_{1}, x_{2}, x_{3}

as well as

y_{1}, y_{2}, y_{3}

are in G.P with same common ratio, then the points

P (x_{1}, y_{1})

Q (x_{2}, y_{2})

R (x_{3}, y_{3})

ABCD is a parallelogram with vertices A (X₁, Y₁) , B(X₂, Y₂) and C (X₃, Y₃). Then the coordinates of the fourth vertex D in terms of the coordinates of A, B and C are

Q. If

x_{1}, x_{2}, x_{3}

as well as

y_{1}, y_{2}, y_{3}

are in G.P with same common ratio, then the point

P (x_{1}, y_{1}), Q (x_{2}, y_{2})

and

R (x_{3}, y_{3})

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The co-ordinates of centroid of a triangle having vertices P(x1,y1),Q(x2,y2) and R(x3,y3) are

The correct option is C (x1+x2+x33,y1+y1+y33)x=x1+x2+x33,y=y1+y2+y33So, option C is correct.

The co-ordinates of centroid of a triangle having vertices $P (x_{1}, y_{1}), Q (x_{2}, y_{2})$ and $R (x_{3}, y_{3})$ are

The correct option is C $(\begin{matrix} \frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{1} + y_{3}}{3} \end{matrix})$
$x = \frac{x_{1} + x_{2} + x_{3}}{3}, y = \frac{y_{1} + y_{2} + y_{3}}{3}$
So, option C is correct.