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TS EAMCET · Maths · Straight Lines

If \(\mathrm{A}=(0,1), \mathrm{B}=(1,2), \mathrm{C}=(-2,1)\) then the equation of the locus of a point P such that area of triangle \(\mathrm{PAB}=\) area of triangle PAC is

  1. A \(x^2-2 x y-3 y^2+2 x+6 y-3=0\)
  2. B \(x^2+2 x y-3 y^2+2 x+6 y-4=0\)
  3. C \(x^2-2 x y-3 y^2+2 x-6 y+4=0\)
  4. D \(x^2-2 x y+3 y^2-2 x+6 y-3=0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(x^2-2 x y-3 y^2+2 x+6 y-3=0\)

Step-by-step Solution

Detailed explanation

P = \( (x,y) \) Area(\(\Delta PAB\)) = \(\frac{1}{2}|x(2-1) + 1(1-y) + 0(y-2)| = \frac{1}{2}|x+1-y|\) Area(\(\Delta PAC\)) = \(\frac{1}{2}|x(1-1) + (-2)(1-y) + 0(y-1)| = \frac{1}{2}|-2+2y|\) Area(\(\Delta PAB\)) = Area(\(\Delta PAC\)) \(\Rightarrow |x-y+1| = |2y-2|\)…