Question

Find a relation between x and y such that the point P(x, y) is equidistant from the points A(5, 3) and B(1, 7).

Hint:

Geometrically, the set of all points equidistant from two fixed points is the perpendicular bisector of the segment joining those points.
You can verify your linear equation by checking if it passes through the midpoint of \(AB\), which is \((3, 5)\).
Indeed, \(3 - 5 + 2 = 0\), which is correct!

Answer 1

Step 1: Understanding the Question:
We are given a point \(P(x, y)\) which is at an equal distance from \(A(5, 3)\) and \(B(1, 7)\). We need to derive a linear relationship between \(x\) and \(y\).

Step 2: Key Formula or Approach:
Since \(P\) is equidistant from \(A\) and \(B\), we have:
\[ PA = PB \implies PA^2 = PB^2 \]
We use the distance formula:
\[ d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \]

Step 3: Detailed Explanation:

• Write the square of distances \(PA^2\) and \(PB^2\):
\[ PA^2 = (x - 5)^2 + (y - 3)^2 \]
\[ PB^2 = (x - 1)^2 + (y - 7)^2 \]

• Equate the two expressions:
\[ (x - 5)^2 + (y - 3)^2 = (x - 1)^2 + (y - 7)^2 \]

• Expand both sides using the identity \((a - b)^2 = a^2 - 2ab + b^2\):
\[ (x^2 - 10x + 25) + (y^2 - 6y + 9) = (x^2 - 2x + 1) + (y^2 - 14y + 49) \]

• Cancel out the quadratic terms \(x^2\) and \(y^2\) from both sides:
\[ -10x - 6y + 34 = -2x - 14y + 50 \]

• Group all terms on one side of the equation:
\[ -10x + 2x - 6y + 14y + 34 - 50 = 0 \]
\[ -8x + 8y - 16 = 0 \]

• Divide the entire equation by \(-8\) to simplify:
\[ x - y + 2 = 0 \]


Step 4: Final Answer:
The required relation between \(x\) and \(y\) is \(x - y + 2 = 0\).