Answer
Verified
480k+ views
Hint: The coordinates of centroid of a triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ is given as $\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$ .
Statement-1: The given triangle is $\Delta ABC$. We will consider the vertices of $\Delta ABC$ to be given as $A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)\ and C\left( {{x}_{3}},{{y}_{3}} \right)$.
Now , from the diagram , we can see that $P,Q,R$ are the midpoints of sides $AB,AC$ and $BC$ respectively .
Now , we know that the midpoint of line joining $\left( {{a}_{1}},{{b}_{1}} \right)\ and \left( {{a}_{2}},{{b}_{2}} \right)$ is given as
$\left( \dfrac{{{a}_{1}}+{{a}_{2}}}{2},\dfrac{{{b}_{1}}+{{b}_{2}}}{2} \right)$
So , midpoint of $AB$ i.e., $P$ is given as
\[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]
But , in the question , it is given that coordinates of $P$ are $\left( 0,0 \right)$.
So , $\dfrac{{{x}_{1}}+{{x}_{2}}}{2}=0\Rightarrow {{x}_{1}}+{{x}_{2}}=0.........\left( i \right)$
And $\dfrac{{{y}_{1}}+{{y}_{2}}}{2}=0\Rightarrow {{y}_{1}}+{{y}_{2}}=0.........\left( ii \right)$
Again , midpoint of $AC$ is $Q$. So , the coordinates of $Q$ are given as
\[\left( \dfrac{{{x}_{1}}+{{x}_{3}}}{2},\dfrac{{{y}_{1}}+{{y}_{3}}}{2} \right)\]
But in the question, coordinates of $Q$ are given as $\left( 1,2 \right)$.
So , $\begin{align}
& \dfrac{{{x}_{1}}+{{x}_{3}}}{2}=1\Rightarrow {{x}_{1}}+{{x}_{3}}=2......\left( iii \right) \\
& \\
\end{align}$
And $\dfrac{{{y}_{1}}+{{y}_{2}}}{2}=2\Rightarrow {{y}_{1}}+{{y}_{3}}=4........\left( iv \right)$
Again , midpoint of $BC$ is $R$. So , the coordinates of $R$ are given as
\[\left( \dfrac{{{x}_{2}}+{{x}_{3}}}{2},\dfrac{{{y}_{2}}+{{y}_{3}}}{2} \right)\]
But in the question, coordinates of $R$ are given as $\left( -3,4 \right)$.
So, $\dfrac{{{x}_{2}}+{{x}_{3}}}{2}=-3\Rightarrow {{x}_{2}}+{{x}_{3}}=-6..........\left( v \right)$
And $\dfrac{{{y}_{2}}+{{y}_{3}}}{2}=4\Rightarrow {{y}_{2}}+{{y}_{3}}=8..........\left( vi \right)$
Now , we will add the equations $\left( i \right),\left( iii \right)$ and $\left( v \right)$.
On adding equations $\left( i \right),\left( iii \right)$ and $\left( v \right)$ , we get
$\begin{align}
& {{x}_{1}}+{{x}_{2}}+{{x}_{1}}+{{x}_{3}}+{{x}_{2}}+{{x}_{3}}=0+2+\left( -6 \right) \\
& \Rightarrow 2\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)=-4 \\
& \Rightarrow \left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)=-2...........\left( vii \right) \\
\end{align}$
Now , we will add the equations $\left( ii \right),\left( iv \right)$and $\left( vi \right)$.
On adding equations $\left( ii \right),\left( iv \right)$and $\left( vi \right)$, we get
$\begin{align}
& {{y}_{1}}+{{y}_{2}}+{{y}_{1}}+{{y}_{3}}+{{y}_{2}}+{{y}_{3}}=0+4+8 \\
& \Rightarrow 2\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)=12 \\
& \Rightarrow \left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)=6...................\left( viii \right) \\
\end{align}$
Now , we know the coordinates of centroid of a triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ is given as $\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$ .
So , the coordinates of centroid of $\Delta ABC$is
$G\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$
But , from equations $\left( vii \right)$and $\left( viii \right)$, we have
${{x}_{1}}+{{x}_{2}}+{{x}_{3}}=-2$ and ${{\text{y}}_{1}}+{{y}_{2}}+{{y}_{3}}=6\text{ }$
So , coordinates of the centroid of $\Delta ABC$are $G\left( \dfrac{-2}{3},2 \right)$ .
Hence , the statement $\left( 1 \right)$ is true.
Statement $2$: Let the vertices of triangle be $A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)\And C\left( {{x}_{3}},{{y}_{3}} \right)$
So the midpoint of $AB$ is $D\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ , $\text{BC}$ is $\text{E}\left( \dfrac{{{x}_{2}}+{{x}_{3}}}{2},\dfrac{{{y}_{2}}+{{y}_{3}}}{2} \right)$ and $AC$ is $\text{F}\left( \dfrac{{{x}_{1}}+{{x}_{3}}}{2},\dfrac{{{y}_{1}}+{{y}_{3}}}{2} \right)$.
Now , we will find the centroid of $\Delta ABC$.
We know the coordinates of centroid of a triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ is given as $\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$.
So , the centroid of $\Delta ABC$ is \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)\] .
Now , we will find the centroid of $\Delta DEF$.
The centroid of $\Delta DEF$ is given as $\left( \dfrac{\dfrac{{{x}_{1}}+{{x}_{2}}}{2}+\dfrac{{{x}_{2}}+{{x}_{3}}}{2}+\dfrac{{{x}_{3}}+{{x}_{1}}}{2}}{3},\dfrac{\dfrac{{{y}_{1}}+{{y}_{2}}}{2}+\dfrac{{{y}_{3}}+{{y}_{2}}}{2}+\dfrac{{{y}_{1}}+{{y}_{3}}}{2}}{3} \right)$
$=\left( \dfrac{\dfrac{2\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)}{2}}{3},\dfrac{\dfrac{2\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)}{2}}{3} \right)$
$=\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$
= centroid of $\Delta ABC$.
Hence , the statement $\left( 2 \right)$ is true.
So , we can conclude that statement $\left( 1 \right)$ is true, statement $\left( 2 \right)$ is true and statement $\left( 2 \right)$ is a correct explanation for statement$\left( 1 \right)$.
So, (1) Statement $\left( 1 \right)$ is true, statement $\left( 2 \right)$ is true; statement $\left( 2 \right)$ is a correct explanation for statement $\left( 1 \right)$.
Note: The midpoint of line joining the points $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$ is given as:
$\left( \dfrac{\left( {{x}_{1}}+{{x}_{2}} \right)}{2},\dfrac{\left( {{y}_{1}}+{{y}_{2}} \right)}{2} \right)$ and not $\left( \dfrac{\left( {{x}_{1}}-{{x}_{2}} \right)}{2},\dfrac{\left( {{y}_{1}}-{{y}_{2}} \right)}{2} \right)$ . Students often get confused between the two. Due to this confusion , they generally end up getting a wrong answer . So , such mistakes should be avoided .
Statement-1: The given triangle is $\Delta ABC$. We will consider the vertices of $\Delta ABC$ to be given as $A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)\ and C\left( {{x}_{3}},{{y}_{3}} \right)$.
Now , from the diagram , we can see that $P,Q,R$ are the midpoints of sides $AB,AC$ and $BC$ respectively .
Now , we know that the midpoint of line joining $\left( {{a}_{1}},{{b}_{1}} \right)\ and \left( {{a}_{2}},{{b}_{2}} \right)$ is given as
$\left( \dfrac{{{a}_{1}}+{{a}_{2}}}{2},\dfrac{{{b}_{1}}+{{b}_{2}}}{2} \right)$
So , midpoint of $AB$ i.e., $P$ is given as
\[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\]
But , in the question , it is given that coordinates of $P$ are $\left( 0,0 \right)$.
So , $\dfrac{{{x}_{1}}+{{x}_{2}}}{2}=0\Rightarrow {{x}_{1}}+{{x}_{2}}=0.........\left( i \right)$
And $\dfrac{{{y}_{1}}+{{y}_{2}}}{2}=0\Rightarrow {{y}_{1}}+{{y}_{2}}=0.........\left( ii \right)$
Again , midpoint of $AC$ is $Q$. So , the coordinates of $Q$ are given as
\[\left( \dfrac{{{x}_{1}}+{{x}_{3}}}{2},\dfrac{{{y}_{1}}+{{y}_{3}}}{2} \right)\]
But in the question, coordinates of $Q$ are given as $\left( 1,2 \right)$.
So , $\begin{align}
& \dfrac{{{x}_{1}}+{{x}_{3}}}{2}=1\Rightarrow {{x}_{1}}+{{x}_{3}}=2......\left( iii \right) \\
& \\
\end{align}$
And $\dfrac{{{y}_{1}}+{{y}_{2}}}{2}=2\Rightarrow {{y}_{1}}+{{y}_{3}}=4........\left( iv \right)$
Again , midpoint of $BC$ is $R$. So , the coordinates of $R$ are given as
\[\left( \dfrac{{{x}_{2}}+{{x}_{3}}}{2},\dfrac{{{y}_{2}}+{{y}_{3}}}{2} \right)\]
But in the question, coordinates of $R$ are given as $\left( -3,4 \right)$.
So, $\dfrac{{{x}_{2}}+{{x}_{3}}}{2}=-3\Rightarrow {{x}_{2}}+{{x}_{3}}=-6..........\left( v \right)$
And $\dfrac{{{y}_{2}}+{{y}_{3}}}{2}=4\Rightarrow {{y}_{2}}+{{y}_{3}}=8..........\left( vi \right)$
Now , we will add the equations $\left( i \right),\left( iii \right)$ and $\left( v \right)$.
On adding equations $\left( i \right),\left( iii \right)$ and $\left( v \right)$ , we get
$\begin{align}
& {{x}_{1}}+{{x}_{2}}+{{x}_{1}}+{{x}_{3}}+{{x}_{2}}+{{x}_{3}}=0+2+\left( -6 \right) \\
& \Rightarrow 2\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)=-4 \\
& \Rightarrow \left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)=-2...........\left( vii \right) \\
\end{align}$
Now , we will add the equations $\left( ii \right),\left( iv \right)$and $\left( vi \right)$.
On adding equations $\left( ii \right),\left( iv \right)$and $\left( vi \right)$, we get
$\begin{align}
& {{y}_{1}}+{{y}_{2}}+{{y}_{1}}+{{y}_{3}}+{{y}_{2}}+{{y}_{3}}=0+4+8 \\
& \Rightarrow 2\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)=12 \\
& \Rightarrow \left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)=6...................\left( viii \right) \\
\end{align}$
Now , we know the coordinates of centroid of a triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ is given as $\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$ .
So , the coordinates of centroid of $\Delta ABC$is
$G\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$
But , from equations $\left( vii \right)$and $\left( viii \right)$, we have
${{x}_{1}}+{{x}_{2}}+{{x}_{3}}=-2$ and ${{\text{y}}_{1}}+{{y}_{2}}+{{y}_{3}}=6\text{ }$
So , coordinates of the centroid of $\Delta ABC$are $G\left( \dfrac{-2}{3},2 \right)$ .
Hence , the statement $\left( 1 \right)$ is true.
Statement $2$: Let the vertices of triangle be $A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)\And C\left( {{x}_{3}},{{y}_{3}} \right)$
So the midpoint of $AB$ is $D\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ , $\text{BC}$ is $\text{E}\left( \dfrac{{{x}_{2}}+{{x}_{3}}}{2},\dfrac{{{y}_{2}}+{{y}_{3}}}{2} \right)$ and $AC$ is $\text{F}\left( \dfrac{{{x}_{1}}+{{x}_{3}}}{2},\dfrac{{{y}_{1}}+{{y}_{3}}}{2} \right)$.
Now , we will find the centroid of $\Delta ABC$.
We know the coordinates of centroid of a triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ is given as $\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$.
So , the centroid of $\Delta ABC$ is \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)\] .
Now , we will find the centroid of $\Delta DEF$.
The centroid of $\Delta DEF$ is given as $\left( \dfrac{\dfrac{{{x}_{1}}+{{x}_{2}}}{2}+\dfrac{{{x}_{2}}+{{x}_{3}}}{2}+\dfrac{{{x}_{3}}+{{x}_{1}}}{2}}{3},\dfrac{\dfrac{{{y}_{1}}+{{y}_{2}}}{2}+\dfrac{{{y}_{3}}+{{y}_{2}}}{2}+\dfrac{{{y}_{1}}+{{y}_{3}}}{2}}{3} \right)$
$=\left( \dfrac{\dfrac{2\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)}{2}}{3},\dfrac{\dfrac{2\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)}{2}}{3} \right)$
$=\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$
= centroid of $\Delta ABC$.
Hence , the statement $\left( 2 \right)$ is true.
So , we can conclude that statement $\left( 1 \right)$ is true, statement $\left( 2 \right)$ is true and statement $\left( 2 \right)$ is a correct explanation for statement$\left( 1 \right)$.
So, (1) Statement $\left( 1 \right)$ is true, statement $\left( 2 \right)$ is true; statement $\left( 2 \right)$ is a correct explanation for statement $\left( 1 \right)$.
Note: The midpoint of line joining the points $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$ is given as:
$\left( \dfrac{\left( {{x}_{1}}+{{x}_{2}} \right)}{2},\dfrac{\left( {{y}_{1}}+{{y}_{2}} \right)}{2} \right)$ and not $\left( \dfrac{\left( {{x}_{1}}-{{x}_{2}} \right)}{2},\dfrac{\left( {{y}_{1}}-{{y}_{2}} \right)}{2} \right)$ . Students often get confused between the two. Due to this confusion , they generally end up getting a wrong answer . So , such mistakes should be avoided .
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE