Question

If the ratio of the distances of a variable point \(P\) from the point \((1,1)\) and the line \(x - y + 2 = 0\) is \(1/\sqrt{2}\), then the equation of the locus of \(P\) is:

• A. \(x^2 + 2xy + y^2 - 8x = 0\)
• B. \(3x^2 + 2xy + 3y^2 - 12x - 4y + 4 = 0\)
• C. \(x^2 + 2xy + y^2 - 12x + 4y + 4 = 0\)
• D. \(x^2 + 2xy + y^2 - 8x + 8y = 0\)

Hint:

For the distance from a point to a line, use the formula \(d = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}\).
- Use the given ratio condition to set up an equation and simplify to find the locus equation.

Answer 1

The Correct Option is B

Let the coordinates of the point \(P\) be \((x,y)\). The distance of \(P\) from the point \((1,1)\) is given by: \[ d_1 = \sqrt{(x-1)^2 + (y-1)^2}. \] The distance of \(P\) from the line \(x - y + 2 = 0\) is given by the formula for the distance from a point to a line: \[ d_2 = \frac{|x - y + 2|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y + 2|}{\sqrt{2}}. \] According to the given condition, the ratio of these distances is: \[ \frac{d_1}{d_2} = \frac{1}{\sqrt{2}}. \] Thus, \[ \frac{\sqrt{(x-1)^2 + (y-1)^2}}{\frac{|x - y + 2|}{\sqrt{2}}} = \frac{1}{\sqrt{2}}. \] Simplifying: \[ \sqrt{2} \sqrt{(x-1)^2 + (y-1)^2} = |x - y + 2|. \] Squaring both sides: \[ 2((x-1)^2 + (y-1)^2) = (x - y + 2)^2. \] Expanding both sides: \[ 2(x^2 - 2x + 1 + y^2 - 2y + 1) = x^2 - 2xy + y^2 + 4x - 4y + 4. \] Simplifying: \[ 2x^2 - 4x + 2 + 2y^2 - 4y + 2 = x^2 - 2xy + y^2 + 4x - 4y + 4. \] Collecting like terms: \[ x^2 + 2xy + 3y^2 - 12x - 4y + 4 = 0. \] Thus, the equation of the locus of \(P\) is \(\boxed{3x^2 + 2xy + 3y^2 - 12x - 4y + 4 = 0}\).