The equation of the locus of all points equid | vedclass

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Question

(A) Let the point be $P(x, y)$.The distance from $P(x, y)$ to the point $(4, 2)$ is $\sqrt{(x - 4)^2 + (y - 2)^2}$.The distance from $P(x, y)$ to the

Vedclass · Accepted Answer

$x^2 - 8x - 4y + 20 = 0$

Vedclass · Answer

(A) Let the point be $P(x, y)$.The distance from $P(x, y)$ to the point $(4, 2)$ is $\sqrt{(x - 4)^2 + (y - 2)^2}$.The distance from $P(x, y)$ to the $x$-axis is $|y|$.Since the point is equidistant,we have $\sqrt{(x - 4)^2 + (y - 2)^2} = |y|$.Squaring both sides,we get $(x - 4)^2 + (y - 2)^2 = y^2$.Expanding the terms: $x^2 - 8x + 16 + y^2 - 4y + 4 = y^2$.Simplifying,we get $x^2 - 8x - 4y + 20 = 0$.

The equation of the locus of all points equidistant from the point $(4, 2)$ and the $x$-axis is:

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