Question

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

• A. \( x^2 - 2xy - 3y^2 + 2x + 6y - 3 = 0 \)
• B. \( x^2 + 2xy - 3y^2 + 2x + 6y - 4 = 0 \)
• C. \( x^2 - 2xy - 3y^2 + 2x - 6y + 4 = 0 \)
• D. \( x^2 - 2xy + 3y^2 - 2x + 6y - 3 = 0 \)

Hint:

When dealing with locus problems involving areas, use the determinant formula for the area of a triangle. Squaring both sides of an equation involving absolute values is a standard technique to eliminate them and proceed with algebraic manipulation.

Answer 1

The Correct Option is A

Let the coordinates of the point P be (x, y).
The area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is given by \( \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| \).
Area(PAB) = \( \frac{1}{2} |x(2-1) + 1(1-y) + 0(y-2)| = \frac{1}{2} |x+1-y| \).
Area(PAC) = \( \frac{1}{2} |x(1-1) - 2(1-y) + 0(y-1)| = \frac{1}{2} |-2+2y| = |y-1| \).
Given that Area(PAB) = Area(PAC).
\( \frac{1}{2} |x-y+1| = |y-1| \implies |x-y+1| = 2|y-1| \).
Squaring both sides to remove the absolute values:
\( (x-y+1)^2 = (2(y-1))^2 = 4(y-1)^2 \).
\( x^2 + y^2 + 1 - 2xy + 2x - 2y = 4(y^2 - 2y + 1) \).
\( x^2 + y^2 - 2xy + 2x - 2y + 1 = 4y^2 - 8y + 4 \).
Rearranging all terms to one side gives the locus:
\( x^2 - 3y^2 - 2xy + 2x + 6y - 3 = 0 \).
This matches option (A).