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Question

Mathematics Question on introduction to three dimensional geometry

The point on the x-axis equidistant from the points (4,3,1)(4,3,1) and (−2, −6, −2)(-2,\,-6,\,-2) is

A

(0, −1,0)(0,\,-1,0)

B

(0, 1,−6)(0,\,1,-6)

C

(−32,0,0)\left( -\frac{3}{2},0,0 \right)

D

(−12,0,0)\left( -\frac{1}{2},0,0 \right)

Answer

(−32,0,0)\left( -\frac{3}{2},0,0 \right)

Explanation

Solution

Let the point on x-axis is (x,0,0)(x,0,0)
According to the question,
(x−4)2+(0−3)2+(0−1)2\sqrt{{{(x-4)}^{2}}+{{(0-3)}^{2}}+{{(0-1)}^{2}}}
=(x+2)2+(0+6)2+(0+2)2=\sqrt{{{(x+2)}^{2}}+{{(0+6)}^{2}}+{{(0+2)}^{2}}}
⇒\Rightarrow x2+16−8x+9+1=x2+4+4x+36+4{{x}^{2}}+16-8x+9+1={{x}^{2}}+4+4x+36+4
⇒\Rightarrow 12x=26−44=−1812x=26-44=-18
⇒\Rightarrow x=−1812=−32x=\frac{-18}{12}=\frac{-3}{2}
Hence, required point is (−32,0,0)\left( \frac{-3}{2},0,0 \right) .