Prove that 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.
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
Let P(x, y) be the point equidistant from A(7, 1) and B(3, 5).
Given: PA = PB
$$\sqrt{(x-7)^2+(y-1)^2} = \sqrt{(x-3)^2+(y-5)^2}$$
Squaring both sides:
$$(x-7)^2+(y-1)^2 = (x-3)^2+(y-5)^2$$
$$x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25$$
$$-14x - 2y + 50 = -6x - 10y + 34$$
$$-14x + 6x - 2y + 10y = 34 - 50$$
$$-8x + 8y = -16$$
$$x - y = 2$$
$$\boxed{x = y + 2}$$
Hence, the abscissa (x) is 2 more than the ordinate (y). Proved.
Source: Chapter 7, Section 7.2 (Distance Formula)
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Explanation
- The key step is setting PA = PB using the distance formula, then squaring to remove the square roots.
- After expanding and simplifying, all $x^2$ and $y^2$ terms cancel, leaving a linear relation.
- The result $x = y + 2$ directly means abscissa = ordinate + 2, which is exactly what had to be proved.
- Examiners award 1 mark for correctly setting up PA = PB and squaring, and 1 mark for the correct simplification leading to $x - y = 2$.