Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).
Step-by-Step Solution
Key Concept: Use the distance formula to set the distances from (x, y) to the two given points equal. Squaring eliminates the square‑roots and yields a linear equation, which is the required relation between x and y. This relation represents the perpendicular bisector of the line segment joining the two fixed points.
1. Write the distances using the distance formula
\[
\text{Distance to }(3,6) = \sqrt{(x-3)^2+(y-6)^2},\qquad
\text{Distance to }(-3,4) = \sqrt{(x+3)^2+(y-4)^2}.
\]
2. Equate the two distances (since the point is equidistant):
\[
\sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x+3)^2+(y-4)^2}.
\]
3. Square both sides to remove the square‑roots:
\[
(x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2.
\]
4. Expand each side
\[
\begin{aligned}
\text{Left side}&: (x^2-6x+9)+(y^2-12y+36) \
&= x^2+y^2-6x-12y+45,\\[4pt]
\text{Right side}&: (x^2+6x+9)+(y^2-8y+16) \
&= x^2+y^2+6x-8y+25.
\end{aligned}
\]
5. Cancel the common terms $x^2$ and $y^2$ and bring all terms to one side:
\[
-6x-12y+45 = 6x-8y+25 \quad\Rightarrow\quad -6x-12y+45-6x+8y-25 =0.
\]
6. Simplify
\[
-12x-4y+20 =0 \quad\Rightarrow\quad 12x+4y-20 =0.
\]
7. Divide by 4 to obtain the simplest linear relation:
\[
3x + y - 5 = 0 \quad\text{or}\quad y = 5-3x.
\]
Thus the required relation between $x$ and $y$ is $3x + y = 5$.
Correct Answer: 3x + y = 5