Q1. [2] § 7.1 Introduction § 7.3 Section Formula § 7.4 Summary
Find a relation between $x$ and $y$ such that the point $P(x, y)$ is equidistant from the points $A(7, 1)$ and $B(3, 5)$.
Previously asked in CBSE board exam
2024 30/3/1 Q23(a) (OR-1)
Generated by claude-sonnet-4-6 · 2026-06-15 10:28 · grounding rag
Model Answer
Given: PA = PB
$$PA = \sqrt{(x-7)^2 + (y-1)^2}, \quad PB = \sqrt{(x-3)^2 + (y-5)^2}$$
Setting PA = PB and squaring both sides:
$$(x-7)^2 + (y-1)^2 = (x-3)^2 + (y-5)^2$$
$$x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25$$
$$-14x - 2y + 50 = -6x - 10y + 34$$
$$-8x + 8y + 16 = 0$$
$$\boxed{x - y = 2}$$
Source: Chapter 7, Exercise 7.1
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Explanation
- Use the distance formula and set PA = PB, then square both sides to remove square roots.
- Expand, cancel $x^2$ and $y^2$ terms, then simplify to get a linear relation.
- Examiners expect the final simplified form $x - y = 2$ (or equivalent). Marks are awarded for correct application of the distance formula, proper squaring/expansion, and the simplified relation.
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