Question

find the equation of the set of points which are equidistant from the points (1,2,3) and (3,2,-1)

Answer 1

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

T(x,y,z)

This is the point equidistant from points P(1,2,3) and Q(3,2,-1)

$\textbf{\underline{Therefore, }}$

We have

TP = TQ

Hence,

TP² = TQ²

(x - 1)² + (y - 2)² + (z - 3)² = (x - 3)² + (y - 2)² + (z + 1)²

Using Identity

${\boxed{\sf\:{(a-b)^2=a^2+b^2-2ab}}}$

${\boxed{\sf\:{(a+b)^2=a^2+b^2+2ab}}}$

Hence,

x² - 2x + 1 + y² - 4y + 4 + z² - 6z + 9 = x² - 6x + 9 + y² - 4y + 4 + z² + 2z + 1

⇒ -2x - 4y - 6z + 14 = -6x - 4y + 2z + 14

⇒ -2x - 6z + 6x - 2z = 0

⇒ 4x - 8z = 0

⇒ 4(x - 2z) = 0

⇒ x - 2z = 0

$\textbf{\underline{Hence\;we\;get\;the\;equation, }}$

$\Large{\boxed{\sf\:{x-2z=0}}$