If a vertex of a triangle is $\left( {{\text{1,1}}} \right)$ and the midpoints of two sides through this vertex are $\left( { - 1,2} \right)$ and $\left( {{\text{3,2}}} \right)$, then the centroid of the triangle is

${\text{A}}{\text{. }}\left( { - 1,\dfrac{7}{3}} \right) $

${\text{B}}{\text{. }}\left( { - \dfrac{1}{3},\dfrac{7}{3}} \right) $

${\text{C}}{\text{. }}\left( {1,\dfrac{7}{3}} \right)$

${\text{D}}{\text{. }}\left( {\dfrac{1}{3},\dfrac{7}{3}} \right)$

Answer

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Hint- Here, we will be proceeding with the help of Midpoint Theorem and formula for coordinates of the centroid of a triangle.

Given, vertex A of the triangle is \[{\text{A}}\left( {1,1} \right)\]

Midpoint of side AB is \[{{\text{M}}_1}\left( { - 1,2} \right)\] and midpoint of side AC is \[{{\text{M}}_2}{\text{(3,2)}}\]

Let the coordinates of vertex B of the triangle be \[\left( {{x_1},{y_1}} \right)\] and that of vertex C be \[\left( {{x_2},{y_2}} \right)\]

As we know that according to Midpoint theorem, coordinates of the midpoint of a line with endpoints as \[\left( {a,b} \right)\] and \[\left( {c,d} \right)\] is given by \[{\text{M}}\left( {\dfrac{{a + c}}{2},\dfrac{{b + d}}{2}} \right)\]

Midpoint of line AB is given by \[{{\text{M}}_1}\left( {\dfrac{{1 + {x_1}}}{2},\dfrac{{1 + {y_1}}}{2}} \right) = \left( { - 1,2} \right)\]

By comparing given midpoint coordinates with those obtained through Midpoint Theorem, we get

\[ \Rightarrow \dfrac{{1 + {x_1}}}{2} = - 1 \Rightarrow 1 + {x_1} = - 2 \Rightarrow {x_1} = - 3\] and \[ \Rightarrow \dfrac{{1 + {y_1}}}{2} = 2 \Rightarrow 1 + {y_1} = 4 \Rightarrow {y_1} = 3\]

Therefore, coordinates of vertex B of the triangle is \[{\text{B}}\left( { - 3,3} \right)\]

Midpoint of line AC is given by \[{{\text{M}}_1}\left( {\dfrac{{1 + {x_2}}}{2},\dfrac{{1 + {y_2}}}{2}} \right) = \left( {3,2} \right)\]

By comparing given midpoint coordinates with those obtained through Midpoint Theorem, we get

\[ \Rightarrow \dfrac{{1 + {x_2}}}{2} = 3 \Rightarrow 1 + {x_2} = 6 \Rightarrow {x_2} = 5\] and \[ \Rightarrow \dfrac{{1 + {y_2}}}{2} = 2 \Rightarrow 1 + {y_2} = 4 \Rightarrow {y_2} = 3\] $$ $$

Therefore, coordinates of vertex C of the triangle is \[{\text{C}}\left( {5,3} \right)\]

Also, we know that coordinates of the centroid of the triangle with vertices \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)\] and \[\left( {{x_3},{y_3}} \right)\] is given by \[\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\]

Therefore, coordinates of the centroid of \[\Delta {\text{ABC}}\] whose vertices are \[{\text{A}}\left( {1,1} \right),{\text{B}}\left( { - 3,3} \right)\] and \[{\text{C}}\left( {5,3} \right)\] is given by \[\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right) = \left( {\dfrac{{1 - 3 + 5}}{3},\dfrac{{1 + 3 + 3}}{3}} \right) = \left( {\dfrac{3}{3},\dfrac{7}{3}} \right) = \left( {1,\dfrac{7}{3}} \right)\]

Note- These types of problems are generally solved by firstly finding out all the vertices of the triangle with the help of midpoint theorem and then using the formula for finding the centroid of the triangle which requires the coordinates of all three vertices of the triangle.

Given, vertex A of the triangle is \[{\text{A}}\left( {1,1} \right)\]

Midpoint of side AB is \[{{\text{M}}_1}\left( { - 1,2} \right)\] and midpoint of side AC is \[{{\text{M}}_2}{\text{(3,2)}}\]

Let the coordinates of vertex B of the triangle be \[\left( {{x_1},{y_1}} \right)\] and that of vertex C be \[\left( {{x_2},{y_2}} \right)\]

As we know that according to Midpoint theorem, coordinates of the midpoint of a line with endpoints as \[\left( {a,b} \right)\] and \[\left( {c,d} \right)\] is given by \[{\text{M}}\left( {\dfrac{{a + c}}{2},\dfrac{{b + d}}{2}} \right)\]

Midpoint of line AB is given by \[{{\text{M}}_1}\left( {\dfrac{{1 + {x_1}}}{2},\dfrac{{1 + {y_1}}}{2}} \right) = \left( { - 1,2} \right)\]

By comparing given midpoint coordinates with those obtained through Midpoint Theorem, we get

\[ \Rightarrow \dfrac{{1 + {x_1}}}{2} = - 1 \Rightarrow 1 + {x_1} = - 2 \Rightarrow {x_1} = - 3\] and \[ \Rightarrow \dfrac{{1 + {y_1}}}{2} = 2 \Rightarrow 1 + {y_1} = 4 \Rightarrow {y_1} = 3\]

Therefore, coordinates of vertex B of the triangle is \[{\text{B}}\left( { - 3,3} \right)\]

Midpoint of line AC is given by \[{{\text{M}}_1}\left( {\dfrac{{1 + {x_2}}}{2},\dfrac{{1 + {y_2}}}{2}} \right) = \left( {3,2} \right)\]

By comparing given midpoint coordinates with those obtained through Midpoint Theorem, we get

\[ \Rightarrow \dfrac{{1 + {x_2}}}{2} = 3 \Rightarrow 1 + {x_2} = 6 \Rightarrow {x_2} = 5\] and \[ \Rightarrow \dfrac{{1 + {y_2}}}{2} = 2 \Rightarrow 1 + {y_2} = 4 \Rightarrow {y_2} = 3\] $$ $$

Therefore, coordinates of vertex C of the triangle is \[{\text{C}}\left( {5,3} \right)\]

Also, we know that coordinates of the centroid of the triangle with vertices \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)\] and \[\left( {{x_3},{y_3}} \right)\] is given by \[\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\]

Therefore, coordinates of the centroid of \[\Delta {\text{ABC}}\] whose vertices are \[{\text{A}}\left( {1,1} \right),{\text{B}}\left( { - 3,3} \right)\] and \[{\text{C}}\left( {5,3} \right)\] is given by \[\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right) = \left( {\dfrac{{1 - 3 + 5}}{3},\dfrac{{1 + 3 + 3}}{3}} \right) = \left( {\dfrac{3}{3},\dfrac{7}{3}} \right) = \left( {1,\dfrac{7}{3}} \right)\]

Note- These types of problems are generally solved by firstly finding out all the vertices of the triangle with the help of midpoint theorem and then using the formula for finding the centroid of the triangle which requires the coordinates of all three vertices of the triangle.

Last updated date: 02nd Oct 2023

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