Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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).

#### 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 $\left( 1, \frac{3}{2}, 4 \right)$

Is there an error in this question or solution?

#### APPEARS IN

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