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Hint:- Coordinates of midpoint of a line is \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)\]. If coordinates of

the end points of the line are \[({x_1},{y_1})\] and \[({x_2},{y_2})\].

We are given with the coordinates of midpoints of the sides of the triangle.

Let the coordinates of the vertices of the triangle be,

\[ \Rightarrow \]Vertices of the triangle are \[(a,b),{\text{ }}(c,d)\] and \[(e,f)\].

So, with the property of mid-point of the two given points.

We can write coordinates of mid-points of the sides of the triangle as,

\[ \Rightarrow \]Midpoint of the sides will be \[\left( {\dfrac{{a + c}}{2},\dfrac{{b + d}}{2}} \right),{\text{ }}\left( {\dfrac{{c + e}}{2},\dfrac{{d + f}}{2}} \right)\]and \[\left( {\dfrac{{a + e}}{2},\dfrac{{b + f}}{2}} \right).\]

As, we know that coordinates of centroid of the triangle are,

\[ \Rightarrow \]Centroid of the triangle is \[\left( {\dfrac{{a + c + e}}{3},\dfrac{{b + d + f}}{3}} \right)\]

And it can be easily seen that coordinates of the centroid of the triangle,

Can be easily obtained by adding the coordinates of the mid-points of its sides

and then dividing that by 3.

So, coordinates of centroid can be written as,

\[ \Rightarrow \]Centroid \[ \equiv \left( {\dfrac{{\left( {\dfrac{{a + c}}{2}} \right) + \left( {\dfrac{{c + e}}{2}} \right) + \left( {\dfrac{{a + e}}{2}} \right)}}{3},\dfrac{{\left( {\dfrac{{b + d}}{2}} \right) + \left( {\dfrac{{d + f}}{2}} \right) + \left( {\dfrac{{b + f}}{2}} \right)}}{3}} \right)\]

So, putting the values of a, b and c in the above point denoted as centroid. We get,

\[ \Rightarrow \]Centroid \[ \equiv \left( {\dfrac{{11 + 2 + 2}}{3},\dfrac{{9 + 1 - 1}}{3}} \right) \equiv \left( {5,3} \right)\]

\[ \Rightarrow \]Hence, the coordinates of the centroid of the triangle will be \[\left( {5,3} \right)\]

\[ \Rightarrow \]Hence, the correct option will be D.

Note:- Whenever we came up with this type of problem then first, we had to assume the

coordinates of vertices of triangle and then find mid-pints in terms of coordinates of

vertices. After that put coordinates of midpoints in terms of vertices of triangle in the formula

centroid triangle.

the end points of the line are \[({x_1},{y_1})\] and \[({x_2},{y_2})\].

We are given with the coordinates of midpoints of the sides of the triangle.

Let the coordinates of the vertices of the triangle be,

\[ \Rightarrow \]Vertices of the triangle are \[(a,b),{\text{ }}(c,d)\] and \[(e,f)\].

So, with the property of mid-point of the two given points.

We can write coordinates of mid-points of the sides of the triangle as,

\[ \Rightarrow \]Midpoint of the sides will be \[\left( {\dfrac{{a + c}}{2},\dfrac{{b + d}}{2}} \right),{\text{ }}\left( {\dfrac{{c + e}}{2},\dfrac{{d + f}}{2}} \right)\]and \[\left( {\dfrac{{a + e}}{2},\dfrac{{b + f}}{2}} \right).\]

As, we know that coordinates of centroid of the triangle are,

\[ \Rightarrow \]Centroid of the triangle is \[\left( {\dfrac{{a + c + e}}{3},\dfrac{{b + d + f}}{3}} \right)\]

And it can be easily seen that coordinates of the centroid of the triangle,

Can be easily obtained by adding the coordinates of the mid-points of its sides

and then dividing that by 3.

So, coordinates of centroid can be written as,

\[ \Rightarrow \]Centroid \[ \equiv \left( {\dfrac{{\left( {\dfrac{{a + c}}{2}} \right) + \left( {\dfrac{{c + e}}{2}} \right) + \left( {\dfrac{{a + e}}{2}} \right)}}{3},\dfrac{{\left( {\dfrac{{b + d}}{2}} \right) + \left( {\dfrac{{d + f}}{2}} \right) + \left( {\dfrac{{b + f}}{2}} \right)}}{3}} \right)\]

So, putting the values of a, b and c in the above point denoted as centroid. We get,

\[ \Rightarrow \]Centroid \[ \equiv \left( {\dfrac{{11 + 2 + 2}}{3},\dfrac{{9 + 1 - 1}}{3}} \right) \equiv \left( {5,3} \right)\]

\[ \Rightarrow \]Hence, the coordinates of the centroid of the triangle will be \[\left( {5,3} \right)\]

\[ \Rightarrow \]Hence, the correct option will be D.

Note:- Whenever we came up with this type of problem then first, we had to assume the

coordinates of vertices of triangle and then find mid-pints in terms of coordinates of

vertices. After that put coordinates of midpoints in terms of vertices of triangle in the formula

centroid triangle.

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