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) $
${\text{C}}{\text{. }}\left( {1,\dfrac{7}{3}} \right)$
${\text{D}}{\text{. }}\left( {\dfrac{1}{3},\dfrac{7}{3}} \right)$
Last updated date: 20th Mar 2023
•
Total views: 308.4k
•
Views today: 2.86k
Answer
308.4k+ views
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.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
