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+9-6 x+y^2+36-12 y=x^2+9+6 x+y^2+16-8 y $
$36-16=6 x+6 x+12 y-8 y $
$20=12 x+4 y $
$3 x+y=5$
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+9-6 x+y^2+36-12 y=x^2+9+6 x+y^2+16-8 y $
$36-16=6 x+6 x+12 y-8 y $
$20=12 x+4 y $
$3 x+y=5$
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