TS EAMCET · Maths · Straight Lines
The equation of the locus of a point which is at a distance of 5 units from a fixed point \((1,4)\) and also from a fixed line \(2 x+3 y-1=0\) is
- A \(9 x^2+12 x y+4 y^2-30 x-108 y+222=0\)
- B \(9 x^2-12 x y+4 y^2-30 x-98 y+220=0\)
- C \(9 x^2+12 x y+4 y^2-22 x-108 y+222=0\)
- D \(9 x^2-12 x y+4 y^2-22 x-98 y+220=0\)
Answer & Solution
Correct Answer
(D) \(9 x^2-12 x y+4 y^2-22 x-98 y+220=0\)
Step-by-step Solution
Detailed explanation
\( \sqrt{(x-1)^2+(y-4)^2} = \frac{|2x+3y-1|}{\sqrt{2^2+3^2}} \) \( (x-1)^2+(y-4)^2 = \frac{(2x+3y-1)^2}{13} \) \( 13(x^2-2x+1+y^2-8y+16) = 4x^2+9y^2+1+12xy-4x-6y \) \( 13x^2+13y^2-26x-104y+221 = 4x^2+9y^2+12xy-4x-6y+1 \) \( 9x^2 - 12xy + 4y^2 - 22x - 98y + 220 = 0 \)
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