Question 1184205
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Find the locus of R if points P(2, -3) and Q(-2, 1) are vertices of triangle PQR and centroid of PQR lies on line 2x + 3y = 1.
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<pre>
Let R = (x,y).


Then the centroid of the triangle PQR  has x-coordinate  {{{x[c]}}} = 1/3 of the sum x-coordinates of the vertices = {{{(2 + (-2) + x)/3}}} = {{{x/3}}}.

                                           y-coordinate  {{{y[c]}}} = 1/3 of the sum y-coordinates of the vertices = {{{((-3) + 1 + y)/3}}} = {{{(y-2)/3}}}.


We want the centroid points  ({{{x[c]}}},{{{y[c]}}}) lie on the line  2x + 3y = 1.


So, we substitute  {{{x/3}}}  for  {{{x[c]}}}  and  {{{(y-2)/3}}}  for  {{{y[c]}}}  into this equation.  We get then

    {{{2*(x/3)}}} + {{{3*((y-2)/3)}}} = 1.


It is EQUIVALENT to 

    {{{2*(x/3)}}} + (y-2) = 1,

    2x + 3*(y-2) = 3,

    2x + 3y - 6 = 3

    2x + 3y = 9.


Thus the locus of points R(x,y) is the straight line  2x + 3y = 9.    <U>ANSWER</U>
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Solved and explained.