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Find the Point on X-axis Which is Equidistant from the Points a (3, 2, 2) and B (5, 5, 4). - Mathematics

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Question

Find the point on x-axis which is equidistant from the points A (3, 2, 2) and B (5, 5, 4).

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Solution

We know that the y and z coordinates of the point on the x-axis are 0.
So, let the required point be C (xyz)
Now, CA = CB

\[\sqrt{\left( 3 - x \right)^2 + \left( 2 - 0 \right)^2 + \left( 2 - 0 \right)^2} = \sqrt{\left( 5 - x \right)^2 + \left( 5 - 0 \right)^2 + \left( 4 - 0 \right)^2}\]
\[ \Rightarrow 9 - 6x + x^2 + 4 + 4 = 25 - 10x + x^2 + 25 + 16\]
\[ \Rightarrow 17 - 6x + x^2 = 66 - 10x + x^2 \]
\[ \Rightarrow 4x = 49\]
\[ \Rightarrow x = \frac{49}{4}\]

Hence, the required point is\[\left( \frac{49}{4}, 0, 0 \right)\]

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Chapter 28: Introduction to three dimensional coordinate geometry - Exercise 28.4 [Page 22]

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RD Sharma Mathematics [English] Class 11
Chapter 28 Introduction to three dimensional coordinate geometry
Exercise 28.4 | Q 11 | Page 22

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