Question

3. Find the coordinates of the point which is equidistant from the vertices of a AAB
where A(3,-1), B (-1,-6) and C(4, -1).​

Answer 1

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 y z− plane is zero.  

Any point in the y z− 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

​  

).