Question

plz answer my question ​

Answer 1

Step-by-step explanation:

$\tt{Let\,\,the\,\,coordinates\,\,of\,\,C\,\,be\,(h,k)}$

Now,

$\sf{Coordintes \: \: of \: \: centroid \: \: be \: \: ( \alpha,\beta )}$

So,

$\sf{ \: \: \: \: \alpha = \dfrac{2 + ( - 2) + h }{3}} \\ \sf{ \beta = \dfrac{- 3 + 1 + k}{3} }$

$\implies \sf{ \alpha = \dfrac{ h }{3}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ \beta = \dfrac{- 2 + k}{3} }$

Now, $(\alpha,\beta)$ lie on the line $2x+3y=1$

So,

$\sf{ 2 \bigg( \dfrac{h}{3} \bigg) + 3 \bigg( \dfrac{ - 2 + k}{3} \bigg)} = 1$

$\sf{ \implies 2(h ) + 3 ( - 2 + k) = 3}$

$\sf{ \implies 2h - 6+ 3k= 3}$

$\sf{ \implies 2h + 3k= 3 + 6}$

$\sf{ \implies 2h + 3k= 9}$

$\rm{ \purple{Hence ,required \: \: locus \: \: \colon \: \: \color{orange}2x + 3y = 9 } }$