Question

If A(1,0), B(0,-2), C(2,-1) are three fixed points, then the equation of the locus of a point P such that area of \( \triangle \text{PAB} \) is equal to area of \( \triangle \text{PAC} \) is

• A. \( x^2 - 2xy - 2y^2 + 2x - 2y + 1 = 0 \)
• B. \( x^2 - 2xy + 2y^2 - 2x + 2y + 1 = 0 \)
• C. \( x^2 - 2xy - 2x + 2y + 1 = 0 \)
• D. \( x^2 - 2xy + 2x - 2y + 1 = 0 \)

Hint:

The area of a triangle with vertices \( (x_1,y_1), (x_2,y_2), (x_3,y_3) \) is \( \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| \). When \( |A|=|B| \), it implies \( A=B \) or \( A=-B \). The combined equation of two lines \( L_1=0 \) and \( L_2=0 \) is given by \( L_1 L_2 = 0 \).

Answer 1

The Correct Option is C

Let P(x, y).
The given fixed points are A(1,0), B(0,-2), and C(2,-1).
The area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is given by the formula \( \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| \).
First, we calculate the area of \( \triangle \text{PAB} \) using vertices P(x,y), A(1,0), B(0,-2): \[ \text{Area}(\triangle \text{PAB}) = \frac{1}{2} |x(0-(-2)) + 1(-2-y) + 0(y-0)| = \frac{1}{2} |2x - 2 - y| \] Next, we calculate the area of \( \triangle \text{PAC} \) using vertices P(x,y), A(1,0), C(2,-1): \[ \text{Area}(\triangle \text{PAC}) = \frac{1}{2} |x(0-(-1)) + 1(-1-y) + 2(y-0)| = \frac{1}{2} |x - 1 - y + 2y| = \frac{1}{2} |x+y-1| \] According to the problem statement, Area(\( \triangle \text{PAB} \)) = Area(\( \triangle \text{PAC} \)).
\[ \frac{1}{2} |2x-y-2| = \frac{1}{2} |x+y-1| \] Multiplying by 2, we get: \[ |2x-y-2| = |x+y-1| \] This equality of absolute values implies two possible cases: Case 1: \( 2x-y-2 = x+y-1 \) Rearranging the terms, we get \( 2x-x-y-y-2+1 = 0 \), which simplifies to: \[ x - 2y - 1 = 0 \cdots (L_1) \] Case 2: \( 2x-y-2 = -(x+y-1) \) \[ 2x-y-2 = -x-y+1 \] Rearranging the terms, we get \( 2x+x-y+y-2-1 = 0 \), which simplifies to: \[ 3x - 3 = 0 \implies x-1=0 \cdots (L_2) \] The locus of point P consists of these two lines.
The combined equation representing this locus is the product of the linear factors \( L_1 \cdot L_2 = 0 \): \[ (x-2y-1)(x-1) = 0 \] Expanding this product: \[ x(x-1) -2y(x-1) -1(x-1) = 0 \] \[ x^2 - x - 2xy + 2y - x + 1 = 0 \] Combining like terms, we get the final equation of the locus: \[ x^2 - 2xy - 2x + 2y + 1 = 0 \] This equation matches option (3).