Question

If $A\left( {\cos \alpha ,\sin \alpha } \right),B\left( {\sin \alpha , - \cos \alpha } \right),C\left( {1,2} \right)$ are the vertices of a $\Delta ABC$, then as a $\alpha $ varies, Find the locus of its centroid.

Answer 1

Hint: By using the formula of centroid, we will find the centroid of $\Delta ABC$. Then assign $x$ of abscissa of centroid and assign $y$ of ordinate of centroid. Then make an equation that does not contain $\alpha $.

Formula Used:
The centroid of a triangle $ = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$
Here, $({x_1},{\rm{ }}{y_1}),({x_2},{\rm{ }}{y_2}),({x_3},{\rm{ }}{y_3})$are vertices of the triangle.
$\begin{array}{l}{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\\{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\end{array}$

Complete step by step solution:
Let $\Delta ABC$ has the centroid $\left( {x,y} \right)$.
So, we can write $x = \dfrac{{\cos \alpha + \sin \alpha + 1}}{3}$ and $y = \dfrac{{\sin \alpha - \cos \alpha + 2}}{3}$
Simplify the equation$x = \dfrac{{\cos \alpha + \sin \alpha + 1}}{3}$.
$\begin{array}{l}3x = \cos \alpha + \sin \alpha + 1\\3x - 1 = \cos \alpha + \sin \alpha \end{array}$ …$\left( 1 \right)$
Simplify the equation $y = \dfrac{{\sin \alpha - \cos \alpha + 2}}{3}$.
$\begin{array}{l}3y = \sin \alpha - \cos \alpha + 2\\3y - 2 = \sin \alpha - \cos \alpha \end{array}$ dfrac …$\left( 2 \right)$
Square equation $\left( 1 \right)$ and $\left( 2 \right)$, add them up.
${\left( {3x - 1} \right)^2} + {\left( {3y - 2} \right)^2} = {\left( {\cos \alpha + \sin \alpha } \right)^2} + {\left( {\sin \alpha - \cos \alpha } \right)^2}$
Simplify the equation
$\begin{array}{l}9{x^2} - 6x + 1 + 9{y^2} - 12y + 4 = {\sin ^2}\alpha + {\cos ^2}\alpha + 2\cos \alpha \sin \alpha + {\sin ^2}\alpha + {\cos ^2}\alpha - 2\cos \alpha \sin \alpha \\9{x^2} + 9{y^2} - 6x - 12y + 5 = 1 + 1\\9{x^2} + 9{y^2} - 6x - 12y + 5 = 2\\9{x^2} + 9{y^2} - 6x - 12y + 3 = 0\end{array}$
Divide the equation by $3$.
$\begin{array}{l}\dfrac{{9{x^2} + 9{y^2} - 6x - 12y + 3}}{3} = 0\\3{x^2} + 3{y^2} - 2x - 4y + 1 = 0\end{array}$
The locus of centroid of $\Delta ABC$ is $3{x^2} + 3{y^2} - 2x - 4y + 1 = 0$

Note: Very often mistake is done here, ${\left( {\cos \alpha + \sin \alpha } \right)^2} + {\left( {\sin \alpha - \cos \alpha } \right)^2}$ is simplified as ${\left( {\sin \alpha } \right)^2} + {\left( {\cos \alpha } \right)^2} = 1$ which is wrong. We need to expand the brackets.