Let the vertex $$C$$ be $$(h,k),$$ then the

centroid of $$\Delta ABC$$ is $$\left( {{{2 + (- 2) + h} \over 3},{{ - 3 + 1 + k} \over 3}} \right)$$

or $$\left( {{h \over 3},{{ - 2 + k} \over 3}} \right).$$ It lies on $$2x+3y=1$$

$$ \Rightarrow {{2h} \over 3} - 2 + k = 1$$

$$ \Rightarrow 2h + 3k = 9$$

$$ \therefore $$ Locus of $$C$$ is $$2x+3y=9$$

Let the required line be $${x \over a} + {y \over b} = 1.......\left( 1 \right)$$

then $$a+b=-1$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.........\left( 2 \right)$$

$$(1)$$ passes through $$(4,3), $$ $$ \Rightarrow {4 \over a} + {3 \over b} = 1$$

$$ \Rightarrow 4b + 3a = ab\,\,...............\left( 3 \right)$$

Eliminating $$b$$ from $$(2)$$ and $$(3),$$ we get

$${a^2} - 4 = 0 \Rightarrow a = \pm 2 \Rightarrow b = - 3$$

$$1$$

$$\therefore$$ Equation of straight lines are

$${x \over 2} + {y \over { - 3}} = 1$$

or $${x \over { - 2}} + {y \over 1} = 1$$

$${\left( {x - {a_1}} \right)^2} + \left( {y - {b_1}} \right){}^2 = \left( {x - {a_2}} \right){}^2 + {\left( {y - {b_2}} \right)^2}$$

$$\left( {a{}_1 - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y$$

$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ + {1 \over 2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right) = 0$$

$$c = {1 \over 2}\left( {{a_2}^2 + {b_2}^2 - a{{{}_1}^2} - {b_1}^2} \right)$$

Taking co-ordinates as

$$\left( {{x \over r},{y \over r}} \right);\left( {x,y} \right)\,\,\& \,\,\left( {xr,yr} \right)$$

Then slope of line joining

$$\left( {{x \over r},{y \over r}} \right),\left( {x,y} \right) = {{y\left( {1 - {1 \over r}} \right)} \over {x\left( {1 - {1 \over r}} \right)}} = {y \over x}$$

and slope of line joining $$(x,y)$$ and $$(xr, yr)$$

$$ = {{y\left( {r - 1} \right)} \over {x\left( {r - 1} \right)}} = {y \over x}$$

$$\therefore$$ $${m_1} = {m_2}$$

$$ \Rightarrow $$ Points lie on the straight line.