Coordinate Geometry
RD Sharma
CBSE
Grade 10
Question:
If the distance between the points $(4, p)$ and $(1, 0)$ is $5$, then the value of $p$ is:
(a) $\pm 4$
(b) $4$
(c) $-4$
(d) $\pm 3$
Step-by-Step Solution
Key Concept: $\sqrt{(1-4)^2 + (0-p)^2} = 5 \Rightarrow 9 + p^2 = 25 \Rightarrow p^2 = 16 \Rightarrow p = \pm 4$.
$(1 - 4)^2 + p^2 = 25 \Rightarrow 9 + p^2 = 25 \Rightarrow p^2 = 16 \Rightarrow p = \pm 4$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Solving $p = \pm 4$: 1.0 Mark
Correct Answer: $\pm 4$
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