find the points on z axis which is equidistance from poinnt A (1,5,7) and B (5,1,-4) - Maths - Introduction to Three Dimensional Geometry

Question

find the points on z axis which is equidistance from poinnt A
(1,5,7) and B (5,1,-4)

Priyanka Kedia · Accepted Answer

Let the point on z-axis is P(0,0,z), which is equidistant from the points A(1,5,7) and B(5,1,-4)

We know the distance formula between two points is,

$d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$ ...... (1)

So, distance between the point P(0,0,z) and A (1,5,7) is,

$d_{1} = \sqrt{{(1 - 0)}^{2} + {(5 - 0)}^{2} + {(7 - z)}^{2}} \Rightarrow d_{1} = \sqrt{26 + {(7 - z)}^{2}}$

And the distance between the point P(0,0,z) and B (5,1,-4) is,

$d_{2} = \sqrt{{(5 - 0)}^{2} + {(1 - 0)}^{2} + {(- 4 - z)}^{2}} \Rightarrow d_{2} = \sqrt{26 + {(- 4 - z)}^{2}}$

Since P(0,0,z) is equidistant from the other two point.

So,

$d_{1} = d_{2} \Rightarrow \sqrt{26 + {(7 - z)}^{2}} = \sqrt{26 + {(- 4 - z)}^{2}} s q u a r i n g o n b o t h s i d e s, w e h a v e, 26 + {(7 - z)}^{2} = 26 + {(- 4 - z)}^{2} \Rightarrow {(7 - z)}^{2} = {(- 4 - z)}^{2} \Rightarrow 49 + z^{2} - 14 z = 16 + z^{2} + 8 z \Rightarrow 49 - 16 = 8 z + 14 z \Rightarrow 33 = 22 z \Rightarrow z = \frac{3}{2} H e n c e, t h e p o int o n z - a x i s w i l l b e (0, 0, \frac{3}{2}) .$

find the points on z axis which is equidistance from poinnt A (1,5,7) and B (5,1,-4).