Find the point on the $x$-axis which is equidistant from $(2, -5)$ and $(-2, 9)$.
Step-by-Step Solution
Key Concept: Let point be $P(x, 0)$. Set $PA^2 = PB^2$ where $A(2,-5), B(-2,9)$.
$(x - 2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0 - 9)^2$. [0.5 Mark]
$x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$. [0.5 Mark]
$-4x + 29 = 4x + 85 \Rightarrow 8x = -56 \Rightarrow x = -7$. Point is $(-7, 0)$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Setting up distance equality for $P(x, 0)$: 1.0 Mark
Solving for $x = -7$ and stating point $(-7, 0)$: 1.0 Mark
Correct Answer: