Question
Find the equation of the set of points which are equidistance from the points (1, 2, 3) and (3, 2, -1).
✓
Answer
Let a point P(x, y, z) be equidistant from the points A(1, 2, 3) and P(3, 2, -1).
Then, $P A = \sqrt { ( x - 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } + ( z - 3 ) ^ { 2 } }$
[$\because$ distance $= \sqrt { \left( x _ { 1 } - x _ { 2 } \right) ^ { 2 } + \left( y _ { 1 } - y _ { 2 } \right) ^ { 2 } + \left( z _ { 1 } - z _ { 2 } \right) ^ { 2 } } ]$
$= \sqrt { x ^ { 2 } - 2 x + 1 + y ^ { 2 } - 4 y + 4 + z ^ { 2 } - 6 z + 9 }$
and $P B = \sqrt { ( x - 3 ) ^ { 2 } + ( y - 2 ) ^ { 2 } + ( z + 1 ) ^ { 2 } }$
[$\because$ distance $= \sqrt { \left( x _ { 1 } - x _ { 2 } \right) ^ { 2 } + \left( y _ { 1 } - y _ { 2 } \right) ^ { 2 } + \left( z _ { 1 } - z _ { 2 } \right) ^ { 2 } } ]$
$= \sqrt { x ^ { 2 } - 6 x + 9 + y ^ { 2 } - 4 y + 4 + z ^ { 2 } + 2 z + 1 }$
$= \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 6 x - 4 y + 2 z + 14 }$
According to the question, PA = PB
$\therefore \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 2 x - 4 y - 6 z + 14 }$
$= \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 6 x - 4 y + 2 z + 14 }$
On squaring both sides, we get
$x^2 + y^2 + z^2 - 2x - 4y - 6z + 14 = x^2 + y^2 + z^2 - 6x - 4y + 2z + 14$
$\Rightarrow$ 4x - 8z = 0
$\Rightarrow$ x - 2z = 0 [dividing both sides by 4]
Then, $P A = \sqrt { ( x - 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } + ( z - 3 ) ^ { 2 } }$
[$\because$ distance $= \sqrt { \left( x _ { 1 } - x _ { 2 } \right) ^ { 2 } + \left( y _ { 1 } - y _ { 2 } \right) ^ { 2 } + \left( z _ { 1 } - z _ { 2 } \right) ^ { 2 } } ]$
$= \sqrt { x ^ { 2 } - 2 x + 1 + y ^ { 2 } - 4 y + 4 + z ^ { 2 } - 6 z + 9 }$
and $P B = \sqrt { ( x - 3 ) ^ { 2 } + ( y - 2 ) ^ { 2 } + ( z + 1 ) ^ { 2 } }$
[$\because$ distance $= \sqrt { \left( x _ { 1 } - x _ { 2 } \right) ^ { 2 } + \left( y _ { 1 } - y _ { 2 } \right) ^ { 2 } + \left( z _ { 1 } - z _ { 2 } \right) ^ { 2 } } ]$
$= \sqrt { x ^ { 2 } - 6 x + 9 + y ^ { 2 } - 4 y + 4 + z ^ { 2 } + 2 z + 1 }$
$= \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 6 x - 4 y + 2 z + 14 }$
According to the question, PA = PB
$\therefore \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 2 x - 4 y - 6 z + 14 }$
$= \sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 6 x - 4 y + 2 z + 14 }$
On squaring both sides, we get
$x^2 + y^2 + z^2 - 2x - 4y - 6z + 14 = x^2 + y^2 + z^2 - 6x - 4y + 2z + 14$
$\Rightarrow$ 4x - 8z = 0
$\Rightarrow$ x - 2z = 0 [dividing both sides by 4]
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