Question
Find a relation between x and y such that the point $(x, y)$ is equidistant from the points $(3, 6)$ and $(-3, 4).$
✓
Answer
Point $(x, y)$ is equidistant form $(3, 6)$ and $(-3, 4).$
Therefore $\sqrt{(\text{x}-3)^2+(\text{y}-6)^2}=\sqrt{\big(\text{x}-(-3)\big)^2+{\text{(y}-4)^2}}$
$\sqrt{(\text{x}-3)^2+(\text{y}-6)^2}=\sqrt{(\text{x}+3)^2+(\text{y}-4)^2}$
Squaring both the side
$\big(\text{x}-3\big)^2+\big(\text{y}-6\big)^2=\big(\text{x}+3\big)^2+\big(\text{y}-4\big)^2$
$x^2 + 9 - 6x + y^2 + 36 - 12y$
$= x^2 + 9 + 6x + y^2 + 16 - 8y$
$36 - 16 = 6x + 6x + 12y - 8y$
$20 = 12x + 4y$
$3x + y = 5$
Therefore $\sqrt{(\text{x}-3)^2+(\text{y}-6)^2}=\sqrt{\big(\text{x}-(-3)\big)^2+{\text{(y}-4)^2}}$
$\sqrt{(\text{x}-3)^2+(\text{y}-6)^2}=\sqrt{(\text{x}+3)^2+(\text{y}-4)^2}$
Squaring both the side
$\big(\text{x}-3\big)^2+\big(\text{y}-6\big)^2=\big(\text{x}+3\big)^2+\big(\text{y}-4\big)^2$
$x^2 + 9 - 6x + y^2 + 36 - 12y$
$= x^2 + 9 + 6x + y^2 + 16 - 8y$
$36 - 16 = 6x + 6x + 12y - 8y$
$20 = 12x + 4y$
$3x + y = 5$
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