Coordinate Geometry
CBSE 2026 Board Exam Set 1 (Code 30/7/1)
CBSE_BOARD_PYQ_2026_30_7_1
Grade 10
Question:
[Section B]
If $Q(0, 1)$ is equidistant from $P(5, -3)$ and $R(x, 6)$, find the value of $x$.
Step-by-Step Solution
Key Concept: $PQ^2 = QR^2 \Rightarrow (5-0)^2 + (-3-1)^2 = (x-0)^2 + (6-1)^2$.
$PQ^2 = 5^2 + (-4)^2 = 25 + 16 = 41$. [0.5 Mark]
$QR^2 = x^2 + 5^2 = x^2 + 25$. [0.5 Mark]
Equating: $x^2 + 25 = 41 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4$. [1.0 Mark]
Correct Answer: $x = \pm 4$
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