Let $$A\left( {2, - 3} \right)$$ and $$B\left( {-2, 1} \right)$$ be vertices of a triangle $$ABC$$. If the centroid of this triangle moves on the line $$2x + 3y = 1$$, then the locus of the vertex $$C$$ is the line
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 $${x \over { - 2}} + {y \over 1} = 1$$
3
AIEEE 2003
MCQ (Single Correct Answer)
If the equation of the locus of a point equidistant from the point $$\left( {{a_{1,}}{b_1}} \right)$$ and $$\left( {{a_{2,}}{b_2}} \right)$$ is
$$\left( {{a_1} - {b_2}} \right)x + \left( {{a_1} - {b_2}} \right)y + c = 0$$ , then the value of $$'c'$$ is
If $${x_1},{x_2},{x_3}$$ and $${y_1},{y_2},{y_3}$$ are both in G.P. with the same common ratio, then the points $$\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$$ and $$\left( {{x_3},{y_3}} \right)$$