Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the point $(3, 6)$ and $(-3, 4)$.
(D) Let the point be $P(x, y)$,$A(3, 6)$,and $B(-3, 4)$.
Since $P$ is equidistant from $A$ and $B$,we have $PA = PB$.
Using the distance formula,$PA^2 = PB^2$.
$(x - 3)^2 + (y - 6)^2 = (x - (-3))^2 + (y - 4)^2$
$(x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2$
Expanding both sides:
$x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16$
Canceling $x^2$ and $y^2$ from both sides:
$-6x - 12y + 45 = 6x - 8y + 25$
Rearranging the terms:
$45 - 25 = 6x + 6x - 8y + 12y$
$20 = 12x + 4y$
Dividing by $4$:
$5 = 3x + y$
Thus,the relation is $3x + y - 5 = 0$.
Since $P$ is equidistant from $A$ and $B$,we have $PA = PB$.
Using the distance formula,$PA^2 = PB^2$.
$(x - 3)^2 + (y - 6)^2 = (x - (-3))^2 + (y - 4)^2$
$(x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2$
Expanding both sides:
$x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16$
Canceling $x^2$ and $y^2$ from both sides:
$-6x - 12y + 45 = 6x - 8y + 25$
Rearranging the terms:
$45 - 25 = 6x + 6x - 8y + 12y$
$20 = 12x + 4y$
Dividing by $4$:
$5 = 3x + y$
Thus,the relation is $3x + y - 5 = 0$.
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