P (x, y) moves such that the area of the tria | vedclass

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Question

$\Delta P Q R=\Delta P S T$

$P(x, y), Q(a, 2 a), R(-a,-2 a)$

$P(x, y), S(a, 2 a) \quad, T(2 a, 3 a)$

$\frac{1}{2}\left|\begin{array}{ccc}x &a

Vedclass · Accepted Answer

$5x - 3y + a = 0$

Vedclass · Answer

$\Delta P Q R=\Delta P S T$

$P(x, y), Q(a, 2 a), R(-a,-2 a)$

$P(x, y), S(a, 2 a) \quad, T(2 a, 3 a)$

$\frac{1}{2}\left|\begin{array}{ccc}x & y & 1 \ a & 2 a & 1 \ -a & -2 a & 1\end{array}ight|=\frac{1}{2}\left|\begin{array}{cccc}x & y & 1 \ a & 2 a & 1 \ 2 a & 3 a & 1\end{array}ight|$

$\left(2 a x-2 a^{2}-a yight)-\left(-2 a^{2}+29 x-4 yight)=\left(29 x+3 a^{2}+2 a yight)-\left(4 a^{2}-3 a x-a yight)$

$2 a x+3 a^{2}+2 a y=4 a^{2}-3a x-a y$

$5 a x+3 a y=a^{2}$

$5 x+3 y=a$

$P (x, y)$ 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 :

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