Question

Find the locus of the point which is equidistant from the point A(0,2,3)and(2,-2,1)

Answer 1

Therefore the locus of (a,b,c) is

x-4y -z+1=0  

Step-by-step explanation:

Let the point be (a,b,c)

The distance between (a,b,c) and (0,2,3) is = $\sqrt{(a-0)^2+(b-2)^2+(c-3)^2}$

The distance between (a,b,c) and (2,-2,1) is = $\sqrt{(a-2)^2+(b+2)^2+(c-1)^2}$

Since the point (a,b,c) is equidistant from (0,2,3) and (2,-2,1)

$\sqrt{(a-0)^2+(b-2)^2+(c-3)^2}$ = $\sqrt{(a-2)^2+(b+2)^2+(c-1)^2}$

⇒ a² +(b-2)²+(c-3)²= (a-2)²+(b+2)²+(c-1)²

⇒a²+b²-4b+4+c²-6c+9=a²-4a+4+b²+4b+4+c²-2c+1

⇒ 4a-8b-4c+4=0

⇒a-4b-c+1=0

Therefore the locus of (a,b,c) is

x-4y -z+1=0       [ putting a = x, b=y and c=z]

Answer 2

AnswEr:

Let P (x,y,z) be any point which is equidistant from A (0,2,3) and B (2,-2,1)

Then,

$\qquad \sf \: PA = PB \\ \\ \\ \implies \sf \: {PA}^{2} = { PB}^{2} \\ \\ \\ \implies \sf \sqrt{ {(x - 0)^2 + {(y - 2)}^{2} } + {(z - 3)}^{2} } \\ \\ \sf = \sqrt{ {(x - 2)}^{2} + {(y + 2)}^{2} + {(z - 1)}^{2} } \\ \\ \\ \implies \sf \: 4x - 8y - 4z + 4 = 0 \\ \\ \\ \implies \sf \blue{ x - 2y - z + 1 = 0}$

Hence, the required locus is x - 2y - z + 1 = 0.