Question

Prove that the abscissa of a point P which is equidistant from points with coordinates A(7, 1) and B(3, 5) is 2 more than its ordinate.

Hint:

Use the distance formula and square both sides to eliminate radicals and solve algebraically.

Answer 1

Problem:
Let point \( P(x, y) \) be equidistant from points \( A(7, 1) \) and \( B(3, 5) \).
We are to find the relationship between \( x \) and \( y \).

Step 1: Use the definition of equidistant
If \( P \) is equidistant from \( A \) and \( B \), then:
\[ PA = PB \]
Apply the distance formula:
\[ \sqrt{(x - 7)^2 + (y - 1)^2} = \sqrt{(x - 3)^2 + (y - 5)^2} \]
Step 2: Square both sides to eliminate square roots
\[ (x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2 \]
Step 3: Expand both sides
Left-hand side: \[ (x - 7)^2 = x^2 - 14x + 49 \\ (y - 1)^2 = y^2 - 2y + 1 \\ \Rightarrow x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 + y^2 - 14x - 2y + 50 \]
Right-hand side: \[ (x - 3)^2 = x^2 - 6x + 9 \\ (y - 5)^2 = y^2 - 10y + 25 \\ \Rightarrow x^2 - 6x + 9 + y^2 - 10y + 25 = x^2 + y^2 - 6x - 10y + 34 \]
Step 4: Cancel like terms on both sides
Remove \( x^2 \) and \( y^2 \) from both sides:
\[ -14x - 2y + 50 = -6x - 10y + 34 \]
Step 5: Bring all terms to one side
\[ -14x + 6x - 2y + 10y = 34 - 50 \Rightarrow -8x + 8y = -16 \]
Step 6: Simplify
\[ -8x + 8y = -16 \Rightarrow x = y + 2 \]
Final Answer:
The abscissa (x-coordinate) is \( \boxed{2 \text{ more than the ordinate (y-coordinate)}} \), i.e., \( \boxed{x = y + 2} \)