JEE Mains · Maths · STD 11 - 9. straight line
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 \(437\)
Answer & Solution
Correct Answer
(C) \(37\)
Step-by-step Solution
Detailed explanation
\( \text { let } P(x, y) \) \( \frac{(x-2)^2+(y-1)^2}{(x-1)^2+(y-3)^2}=\frac{25}{16} \) \( 9 x^2+9 y^2+14 x-118 y+170=0 \) \( a^2+2 b+3 c+4 d+e \) \( =81+18+0+56-118 \) \( =155-118 \) \( =37\)
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