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  • is a variable triangle with the fixed vertex <img alt="\mathrm{C(1, 2)}

\mathrm{ABC} is a variable triangle with the fixed vertex \mathrm{C(1, 2)} and \mathrm{A, B} having the coordinates \mathrm{(cos\ t, sin\ t), (sin\ t - cos\ t)} respectively where t is a parameter. Find the locus of the Centroid of the \mathrm{\bigtriangleup ABC.}

 

Option: 1

\mathrm{x^{2}+y^{2}-2x-4y+2=0}


Option: 2

\mathrm{2\left ( x^{2}+y^{2} \right )-2x-4y+1=0}


Option: 3

\mathrm{3\left ( x^{2} +y^{2}\right )-2x-4y+1=0}


Option: 4

\mathrm{3\left ( x^{2} +y^{2}\right )+2x+4y-1=0}


Answers (1)

best_answer

 Let \mathrm{G} be the centroid in any position.
\mathrm{(3 \alpha-1)^2+(3 \beta-2)^2=(\cos t+\sin t)^2+(\sin t-\cos t)^2}
Then \mathrm{(\alpha, \beta)=\left(\frac{1+\cos t+\sin t}{3}, \frac{2+\sin t-\cos t}{3}\right)}
\mathrm{\therefore \alpha=\frac{1+\cos t+\sin t}{3}, \beta=\frac{2+\sin t-\cos t}{3} \text { or } 3 \alpha-1=\cos t+\sin t \quad \&\ 3 \beta-2=\sin t-\cos t}
Squaring and adding \mathrm{\Rightarrow 2\left(\cos ^2 t+\sin ^2 t\right)=2}
\mathrm{\therefore } the equation of the locus of the centroid is \mathrm{\left ( 3x-1 \right )^{2} +\left ( 3y-2 \right )^{2}=2}
\mathrm{\therefore 3\left(\mathrm{x}^2+\mathrm{y}^2\right)-2 \mathrm{x}-4 \mathrm{y}+1=0.}

Posted by

manish painkra

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