Question

If the points \[\left( {2a,a} \right)\] , \[(a,2a)\] and \[(a,a)\] enclose a triangle of area 18 square. units, then the centroid of the triangle is equal to -
(A) \[(4,4)\]
(B) \[(8,8)\]
(C) \[( - 4, - 4)\]
(D) \[(4\sqrt 2 ,4\sqrt 2 )\]
(E) \[(6,6)\]

Answer 1

Hint: According to the given, use the given points in the formula to get the area of triangle \[ = \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\] and calculate the value of a. Then, calculate centroid of triangle \[ = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)\left( {\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\] by substituting the calculated values.

Formula used:
Here, we can use the formula to find area of triangle \[ = \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\] and centroid of triangle \[ = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)\left( {\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\]

Complete step-by-step answer:
As we are given with the area of triangle =18 and the points \[\left( {2a,a} \right)\] , \[(a,2a)\] and \[(a,a)\].

So, we will substitute the values as \[({x_1},{y_1}) = \]\[\left( {2a,a} \right)\] , \[({x_2},{y_2}) = \]\[(a,2a)\] , \[({x_3},{y_3}) = \]\[(a,a)\] .
Firstly we will calculate area of triangle to get the value of a by using the formula
Area of triangle \[ = \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\] .
After substituting all the given values we get,
$\Rightarrow$ \[18 = \dfrac{1}{2}\left[ {2a\left( {2a - a} \right) + a(a - a) + a(a - 2a)} \right]\]
Taking 2 on the left hand side we get,
$\Rightarrow$ \[36 = \left[ {2a\left( {2a - a} \right) + a(a - a) + a(a - 2a)} \right]\]
After solving all the brackets we get,
$\Rightarrow$ \[36 = \left[ {2{a^2} + 0 - {a^2}} \right]\]
So, \[{a^2} = 36\]
Hence \[a = 6, - 6\]
Now, we will calculate the centroid of the triangle by using the formula
Centroid of triangle \[ = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3}} \right)\left( {\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)\]
By substituting the given values we get,
\[ \Rightarrow \left( {\dfrac{{2a + a + a}}{3}} \right)\left( {\dfrac{{a + 2a + a}}{3}} \right)\]
On simplifying we get,
\[ \Rightarrow \left( {\dfrac{{4a}}{3}} \right)\left( {\dfrac{{4a}}{3}} \right)\]
Here, we will first put the value of \[a = 6\]
So, we get
\[ \Rightarrow \left( {\dfrac{{24}}{3}} \right)\left( {\dfrac{{24}}{3}} \right)\]
On dividing we get,
Centroid of triangle is \[(8,8)\]
Then, we will put the value of \[a = - 6\]
So, we get
\[ \Rightarrow \left( {\dfrac{{ - 24}}{3}} \right)\left( {\dfrac{{ - 24}}{3}} \right)\]
On dividing we get,
Centroid of the triangle is \[( - 8, - 8)\] which is not given in the option.
So, we will consider Centroid of triangle = \[(8,8)\]
So, the correct option is (B) \[(8,8)\]

Note: To solve these types of questions you can also use the alternative method to calculate the area in the above question is given by, We are using the formula as area of triangle \[ = \dfrac{1}{2}\left| A \right|\] as, \[\left| A \right|\] is given by \[\left| \begin{array}{l}2a{\rm{ a 1}}\\{\rm{a 2a 1}}\\{\rm{a a 1}}\end{array} \right|\]
On substituting the values we get,
\[18 = \dfrac{1}{2}\left| \begin{array}{l}2a{\rm{ a 1}}\\{\rm{a 2a 1}}\\{\rm{a a 1}}\end{array} \right|\]
After taking 2 on the left side we get,
\[36 = \left| \begin{array}{l}2a{\rm{ a 1}}\\{\rm{a 2a 1}}\\{\rm{a a 1}}\end{array} \right|\]
On opening determinant we get,
\[36 = 2a(2a - a) - a(a - a) + 1({a^2} - 2{a^2})\]
After simplifying we get,
\[{a^2} = 36\]
So, \[a = \pm 6\]
Therefore, we can calculate the centroid of the triangle by using the same formula as done in the above answer.