TS EAMCET · Maths · Straight Lines
If the distance from a variable point \(P\) to the point \((4,3)\) is equal to the perpendicular distance from \(P\) to the line \(x+2 y-1=0\), then the equation of the locus of the point \(P\) is
- A \(4 x^2+4 x y+y^2-38 x+26 y+124=0\)
- B \(4 x^2-4 x y+y^2-38 x-26 y+124=0\)
- C \(4 x^2-4 x y+y^2+38 x+26 y+124=0\)
- D \(4 x^2-4 x y+y^2-38 x+26 y+124=0\)
Answer & Solution
Correct Answer
(B) \(4 x^2-4 x y+y^2-38 x-26 y+124=0\)
Step-by-step Solution
Detailed explanation
Let \(P(h, k)\) according to question \((h-4)^2+(k-3)^2=\left(\frac{h+2 k-1}{\sqrt{5}}\right)^2\)…
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