Equation of Locus of Point Equidistant from Point and Line

To find the locus of a point $Q(x, y)$ whose distance from the point $P(3, 4)$ is equal to its perpendicular distance from the line $AB: x + 3y - 3 = 0$, we use the definition of a parabola (focus-directrix property).

1. Distance from point $Q(x, y)$ to point $P(3, 4)$:
   $d_1 = \sqrt{(x - 3)^2 + (y - 4)^2}$

2. Perpendicular distance from point $Q(x, y)$ to the line $x + 3y - 3 = 0$:
   $d_2 = \frac{|x + 3y - 3|}{\sqrt{1^2 + 3^2}} = \frac{|x + 3y - 3|}{\sqrt{10}}$

3. Equating the two distances ($d_1 = d_2$):
   $\sqrt{(x - 3)^2 + (y - 4)^2} = \frac{|x + 3y - 3|}{\sqrt{10}}$

4. Squaring both sides:
   $(x - 3)^2 + (y - 4)^2 = \frac{(x + 3y - 3)^2}{10}$

5. Expanding and simplifying:
   $10 \left[ (x^2 - 6x + 9) + (y^2 - 8y + 16) \right] = (x + 3y - 3)^2$

$10(x^2 + y^2 - 6x - 8y + 25) = x^2 + 9y^2 + 9 + 2(x)(3y) + 2(3y)(-3) + 2(x)(-3)$

$10x^2 + 10y^2 - 60x - 80y + 250 = x^2 + 9y^2 + 6xy - 18y - 6x + 9$

Rearranging all terms to one side:
$(10x^2 - x^2) + (10y^2 - 9y^2) - 6xy + (-60x + 6x) + (-80y + 18y) + (250 - 9) = 0$

$9x^2 + y^2 - 6xy - 54x - 62y + 241 = 0$

This matches the equation given in option A.