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9.Straight Line
hard
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 $\mathrm{a}^2+2 \mathrm{~b}+3 \mathrm{c}+4 \mathrm{~d}+\mathrm{e}$ is equal to:
A
$5$
B
$-27$
C
$37$
D
$437$
(JEE MAIN-2024)
Solution
$ \text { let } P(x, y) $
$ \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$
Standard 11
Mathematics
