• # question_answer Let A (2, - 3) and B (- 2,1) be vertices of a$\Delta ABC$ If the centroid of this triangle moves on the line $2x+3y=1,$then the locus of the vertex C is the line A) $2x+3y=9$        B) $2x-3y=7$ C) $3x+2y=5$        D)        $3x-2y=3$

If$A({{x}_{1}},{{y}_{1}}),B({{x}_{2}},{{y}_{2}}),C({{x}_{3}},{{y}_{3}})$are the vertices of a triangle, then the coordinates of the   centroid   of   a   triangle   are$\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$. Let$(x,\text{ }y)$be the coordinates of vertex C and $({{x}_{1}},\text{ }{{y}_{1}})$be the coordinates of centroid of the triangle. $\therefore$  ${{x}_{1}}=\frac{x+2-2}{3}$and${{y}_{1}}=\frac{y-3+1}{3}$ $\Rightarrow$               ${{x}_{1}}=\frac{x}{3}$and${{y}_{1}}=\frac{y-2}{3}$                    ?.(i) Since, the centroid lies on the line$2x+3y=1$. So,${{x}_{1}}$and${{y}_{1}}$will satisfy the equation of line. $\therefore$  $2{{x}_{1}}+3{{y}_{1}}=1$ $\Rightarrow$$\frac{2x}{3}+\frac{3(y-2)}{3}=1$                          [from Eq. (i)] $\Rightarrow$$2x+3y-6=3$ $\Rightarrow$$2x+3y=9$ This equation is locus of the vertex C.