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If the locus of the point, whose distances from the point $(2,1)$ and $(1,3)$ are in the ratio $5: 4$, is $a x^2+b y^2+c x y+d x+e y+170=0$, then the value of $a^2+2 b+3 c+4 d+e$ is equal to :
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Verified Answer
The correct answer is:
37
let $\mathrm{P}(\mathrm{x}, \mathrm{y})$
$\begin{aligned}
& \frac{(x-2)^2+(y-1)^2}{(x-1)^2+(y-3)^2}=\frac{25}{16} \\
& 9 x^2+9 y^2+14 x-118 y+170=0 \\
& a^2+2 b+3 c+4 d+e \\
& =81+18+0+56-118 \\
& =155-118 \\
& =37
\end{aligned}$
$\begin{aligned}
& \frac{(x-2)^2+(y-1)^2}{(x-1)^2+(y-3)^2}=\frac{25}{16} \\
& 9 x^2+9 y^2+14 x-118 y+170=0 \\
& a^2+2 b+3 c+4 d+e \\
& =81+18+0+56-118 \\
& =155-118 \\
& =37
\end{aligned}$
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