Question

find the points in the yz-plane which is equidistant from 3 points(1,-1,0);(2,1,2)and(3,2,-1)​

Answer 1

Answer:

Given points are A(3,2,−1),B(1,−1,0) and C(2,1,2).

We know that x coordinate of every point in yz− plane is zero.

Any point in the yz− plane is of the form P(0,y,z).

Now, according to question

∣AP∣=∣BP∣=∣CP∣ Given

⇒AP

2

=BP

2

⇒(0−3)

2

+(y−2)

2

+(z+1)

2

=(0−1)

2

+(y+1)

2

+(z−0)

2

⇒3y−z−6=0 (i)

and BP

2

=CP

2

⇒(0−1)

2

+(y+1)

2

+(z−0)

2

=(0−2)

2

+(y−1)

2

+(z−2)

2

⇒4y+4z−7=0 (ii)

On solving, (i) and (ii) we get

y=

16

31

and z=

16

−3

.

Hence, the required point is (0,

16

31

16

−3