GSEB Std 10 (SSC) • 2024 • 3 Marks

Coordinate Geometry: Equidistant Point on X-Axis

Official examination question with verified M1/A1 mark scheme and step-by-step mathematical reasoning.

Problem Statement

Find the point on the $x$-axis which is equidistant from $(2, -5)$ and $(-2, 9)$.

Verified Solution & Marking Scheme

Set Up Coordinates and Distance Formula
Let the point on $x$-axis be $P(x, 0)$. Let $A(2, -5)$ and $B(-2, 9)$. Since $PA = PB \implies PA^2 = PB^2$: $(x - 2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0 - 9)^2$
Expand and Solve for x
$x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$ $-4x + 29 = 4x + 85$ $-8x = 85 - 29 = 56 \implies x = -\frac{56}{8} = -7$
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