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KEAM 2018 Math Paper

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Question : 88 of 120

Marks: +1, -0

The point in the xy-plane which is equidistant from (2, 0, 3), (0, 3, 2) and (0, 0, 1) is

(1, 2, 3)
(–3, 2, 0)
(3, –2, 0)
(3, 2, 0)
(3, 2, 1)

Solution:

Let the points are

A(2,0,3), B(0,3,2)

and

D(0,0,1)

We know that

Z

-coordinate of every point an

x y

-plane is zero so let

p(x, y, 0)

be a point on

x y

-plane such that

P A = P B = P C

.

Now,

P A = P B

\Rightarrow P A^{2} = P B^{2}

\Rightarrow (x-2)^{2} + (y-0)^{2} + (0-3)^{2} = (x-0)^{2} + (y-3)^{2}

+(0-2)

\Rightarrow \;\; 4 x - 6 y = 0 \Rightarrow 2 x - 3 y = 0

....(i)

and,

\;\; P B = P C

\Rightarrow \;\; P B^{2} = P C^{2}

\Rightarrow (x-0)^{2} + (y-3)^{2} + (0-2)^{2} = (x-0)^{2} + (y-0)^{2} + (0-1)^{2}

\Rightarrow \;\; -6 y + 12 = 0

\Rightarrow \;\; y = 2

....(ii)

Putting

y = 2

in Eq. (i), we get

x = 3

Hence, the required point is

(3,2,0)

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