Question

If the distance from a variable point \( P \) to the point \( (4,3) \) is equal to the perpendicular distance from \( P \) to the line \( x + 2y - 1 = 0 \), then the equation of the locus of the point \( P \) is:

• A. \( 4x^2 + 4xy + y^2 - 38x + 26y + 124 = 0 \)
• B. \( 4x^2 - 4xy + y^2 - 38x - 26y + 124 = 0 \)
• C. \( 4x^2 - 4xy + y^2 + 38x + 26y + 124 = 0 \)
• D. \( 4x^2 - 4xy - 38x + 26y + 124 = 0 \) \bigskip

Hint:

To determine the locus of a point satisfying a distance condition, set the relevant distances equal, square the resulting equation to eliminate square roots, and simplify by expanding and combining like terms.

Answer 1

The Correct Option is B

Let \( P(x, y) \) be an arbitrary point. 

Step 1: Compute the distance from \( P(x, y) \) to the point \( (4, 3) \). The distance formula gives: \[ d_1 = \sqrt{(x - 4)^2 + (y - 3)^2}. \] \bigskip Step 2: Compute the perpendicular distance from \( P(x, y) \) to the line \( x + 2y - 1 = 0 \). For a line \( Ax + By + C = 0 \), the perpendicular distance is: \[ d_2 = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}. \] For the given line, \( A = 1 \), \( B = 2 \), and \( C = -1 \), hence: \[ d_2 = \frac{|x + 2y - 1|}{\sqrt{1^2 + 2^2}} = \frac{|x + 2y - 1|}{\sqrt{5}}. \] 

Step 3: Set the distances equal. According to the condition, the distance from \( P \) to \( (4, 3) \) equals its perpendicular distance to the line: \[ \sqrt{(x - 4)^2 + (y - 3)^2} = \frac{|x + 2y - 1|}{\sqrt{5}}. \] 

Step 4: Square both sides to remove the square roots: \[ (x - 4)^2 + (y - 3)^2 = \frac{(x + 2y - 1)^2}{5}. \] Multiply both sides by 5: \[ 5\left[(x - 4)^2 + (y - 3)^2\right] = (x + 2y - 1)^2. \] 

Step 5: Expand both sides. Expanding the left-hand side: \[ 5\left(x^2 - 8x + 16 + y^2 - 6y + 9\right) = 5x^2 - 40x + 80 + 5y^2 - 30y + 45. \] Expanding the right-hand side: \[ (x + 2y - 1)^2 = x^2 + 4xy - 2x + 4y^2 - 4y + 1. \] 

Step 6: Combine like terms. Bring all terms to one side: \[ 5x^2 + 5y^2 - 40x - 30y + 125 - x^2 - 4xy + 2x - 4y^2 + 4y - 1 = 0. \] Simplify: \[ 4x^2 - 4xy + y^2 - 38x - 26y + 124 = 0. \] Thus, the equation of the locus of the point \( P \) is: \[ \boxed{4x^2 - 4xy + y^2 - 38x - 26y + 124 = 0}. \]