Question

find the locus of point which move so that It's equidistant fram (1,2) and (2,-3)
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Answer 1

Answer:

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

Then,PA=PB PA 2

=PB 2(x−0) 2 +(y−2)2 +(z−3)

2 = (x−2) 2 +(y+2) 2 +(z−1) 2

4x−8y−4z+4=0

x−2y−z+1=0

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

Answer 2

Answer:

dist(AP)

=dist(BP)

dist(AP)

=dist(BP)

=>√(x−1)2+(y−2)2

=√(X+2)2+(y+1)2

=

>√(x−1)2+(y−2)2

=√(X+2)2+(y+1)2

Simplify to get the equation.

Simpler way is to imagine the locus

It will be a straight line.

It will be perpendicular to AB.

Slope of AB =( 2+1)/(1+2) = 1

Hence slope of locus of P = -1

This P line passes through mid point of AB .

Mid pt of AB = (-1/2, 1/2)

EQ of locus is

(y - 1/2)/(X+ 1/2) = -1✔️✔️