Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Find the Coordinates of the Point Which is Equidistant from the Four Points O(0, 0, 0), A(2, 0, 0), B(0, 3, 0) and C(0, 0, 8). - Mathematics

Find the coordinates of the point which is equidistant  from the four points O(0, 0, 0), A(2, 0, 0), B(0, 3, 0) and C(0, 0, 8).

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Solution

Let P (xyz) be the required point which is equidistant from the points O(0,0,0), A(2,0,0)
B(0,3,0) and C(0,0,8)

Then,
OP = AP 

\[\Rightarrow O P^2 = A P^2\]

\[\therefore x^2 + y^2 + z^2 = \left( x - 2 \right)^2 + y^2 + z^2 \]
\[ \Rightarrow x^2 = \left( x - 2 \right)^2 \]
\[ \Rightarrow x^2 = x^2 - 4x + 4\]
\[ \Rightarrow 4x = 4\]
\[ \Rightarrow x = \frac{4}{4}\]
\[ \therefore x = 1\]
Similarly, we have:
OP = BP
\[\Rightarrow O P^2 = B P^2 \]
\[\therefore x^2 + y^2 + z^2 = x^2 + \left( y - 3 \right)^2 + z^2 \]
\[ \Rightarrow y^2 = y^2 - 6y + 9\]
\[ \Rightarrow 6y = 9\]
\[ \Rightarrow y = \frac{9}{6}\]
\[ \therefore y = \frac{3}{2}\]
Similarly, we also have:

OP CP

\[\Rightarrow O P^2 = C P^2\]

\[\Rightarrow x^2 + y^2 + z^2 = x^2 + y^2 + \left( z - 8 \right)^2 \]
\[ \Rightarrow z^2 = z^2 - 16z + 64\]
\[ \Rightarrow 16z = 64\]
\[ \Rightarrow z = \frac{64}{16}\]
\[ \therefore z = 4\]

Thus, the required point is P \[\left( 1, \frac{3}{2}, 4 \right)\]

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 28 Introduction to three dimensional coordinate geometry
Exercise 28.2 | Q 15 | Page 10
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