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(D) Let the point $P$ be $(x, y)$.According to the problem,the distance from $P(x, y)$ to $(4, 3)$ is equal to the perpendicular distance from $P(x, y

Vedclass · Accepted Answer

$4x^2 - 4xy + y^2 - 38x + 26y + 124 = 0$

Vedclass · Answer

(D) Let the point $P$ be $(x, y)$.According to the problem,the distance from $P(x, y)$ to $(4, 3)$ is equal to the perpendicular distance from $P(x, y)$ to the line $x + 2y - 1 = 0$.Using the distance formula and the perpendicular distance formula:$\sqrt{(x - 4)^2 + (y - 3)^2} = \frac{|x + 2y - 1|}{\sqrt{1^2 + 2^2}}$Squaring both sides:$(x - 4)^2 + (y - 3)^2 = \frac{(x + 2y - 1)^2}{5}$$5(x^2 - 8x + 16 + y^2 - 6y + 9) = x^2 + 4y^2 + 1 + 4xy - 2x - 4y$$5x^2 - 40x + 80 + 5y^2 - 30y + 45 = x^2 + 4y^2 + 4xy - 2x - 4y + 1$$4x^2 + y^2 - 4xy - 38x - 26y + 124 = 0$

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

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