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: Let $P(x,y), A(7,1), B(3,5)$. Set $PA^2 = PB^2$.
Stepwise Solution:
$(x - 7)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2$. [0.5 Mark]
$x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25$. [0.5 Mark]
$-14x - 2y + 50 = -6x - 10y + 34 \Rightarrow -8x + 8y = -16 \Rightarrow x - y = 2$. [1.0 Mark]
Marking Scheme:
• Equidistant equation setup $PA^2 = PB^2$: 0.5 Mark
• Expanding terms: 0.5 Mark
• Simplifying to linear relation $x - y = 2$: 1.0 Mark
Correct Answer: