First slide

Introduction to 3-D Geometry

Question

The point in $x y -$ plane which is equidistant from the points $A (1, - 1, 0), B (2, 1, 2)$ and $C (3, 2, - 1)$

Easy

A

$(3, 1, 0)$
B

$(- \frac{3}{2}, 1, 0)$
C

$(\frac{3}{2}, 1, 0)$
D

$(\frac{31}{10}, 0, \frac{1}{5})$

Solution

Suppose that $P (x_{1}, y_{1}, 0)$ be any point in the $x y -$ plane and suppose that it is equidistant from the points $A (1, - 1, 0), B (2, 1, 2)$ and $C (3, 2, - 1)$
it gives $P A = P B = P C$
$\Rightarrow {(x - 1)}^{2} + {(y + 1)}^{2} + 0 = {(x - 2)}^{2} + {(y - 1)}^{2} + 4 \Rightarrow 2 x + 4 y = 7$
and
$\Rightarrow {(x - 1)}^{2} + {(y + 1)}^{2} + 2 = {(x - 3)}^{2} + {(y - 2)}^{2} + 1 \Rightarrow 2 x + 3 y = 6$
Solving the above equations $y = 1 . x = \frac{3}{2}$
Therefore, the required point is $(\frac{3}{2}, 1, 0)$

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