A) \[2x+3y=9\]
B) \[2x-3y=7\]
C) \[3x+2y=5\]
D) \[3x-2y=3\]
Correct Answer: A
Solution :
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.
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