Answer

Verified

410.4k+ views

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.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

Using the following information to help you answer class 12 chemistry CBSE

Why should electric field lines never cross each other class 12 physics CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Explain why a There is no atmosphere on the moon b class 11 physics CBSE

Difference Between Plant Cell and Animal Cell

Can anyone list 10 advantages and disadvantages of friction

State and prove Bernoullis theorem class 11 physics CBSE

The ice floats on water because A solid have lesser class 9 chemistry CBSE

State Newtons formula for the velocity of sound in class 11 physics CBSE