Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).
Step-by-Step Solution
Key Concept: A point is equidistant from two given points if and only if it lies on the perpendicular bisector of the line segment joining those points. Using the distance formula, set the distances from (x, y) to each given point equal and simplify to obtain the required relation.
1. Write the distance formula for the point \((x, y)\) to each of the given points:
$$\begin{aligned}
d_1 &= \sqrt{(x-7)^2+(y-1)^2},\\[4pt]
d_2 &= \sqrt{(x-3)^2+(y-5)^2}.
\end{aligned}$$
2. Equate the two distances because the point is equidistant:
$$\sqrt{(x-7)^2+(y-1)^2}=\sqrt{(x-3)^2+(y-5)^2}.$$
3. Square both sides to remove the square‑roots (both sides are non‑negative):
$$(x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2.$$
4. Expand each square:
$$\begin{aligned}
(x^2-14x+49)+(y^2-2y+1) &= (x^2-6x+9)+(y^2-10y+25).
\end{aligned}$$
5. Cancel the common terms \(x^2\) and \(y^2\):
$$-14x+49-2y+1 = -6x+9-10y+25.$$
6. Collect like terms on one side:
\[-14x+6x\] + \[-2y+10y\] + \[50-34\] = 0
$$-8x+8y+16=0.$$
7. Divide by 8 to simplify:
$$-x+y+2=0.$$
8. Write the required relation between \(x\) and \(y\):
$$y = x-2.$$
Correct Answer: y = x - 2