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$P (x, y)$ moves such that the area of the triangle formed by $P, Q (a , 2 a)$ and $R (- a, - 2 a)$ is equal to the area of the triangle formed by $P, S (a, 2 a)\,\,\, \&\,\, \,T (2 a, 3 a)$. The locus of $'P'$ is a straight line given by :
$3x - y = a$
$(A)$ or $(C)$ both
$y = 2ax$
$5x - 3y + a = 0$
Solution
$\Delta P Q R=\Delta P S T$
$P(x, y), Q(a, 2 a), R(-a,-2 a)$
$P(x, y), S(a, 2 a) \quad, T(2 a, 3 a)$
$\frac{1}{2}\left|\begin{array}{ccc}x & y & 1 \\ a & 2 a & 1 \\ -a & -2 a & 1\end{array}\right|=\frac{1}{2}\left|\begin{array}{cccc}x & y & 1 \\ a & 2 a & 1 \\ 2 a & 3 a & 1\end{array}\right|$
$\left(2 a x-2 a^{2}-a y\right)-\left(-2 a^{2}+29 x-4 y\right)=\left(29 x+3 a^{2}+2 a y\right)-\left(4 a^{2}-3 a x-a y\right)$
$2 a x+3 a^{2}+2 a y=4 a^{2}-3a x-a y$
$5 a x+3 a y=a^{2}$
$5 x+3 y=a$