Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, -4)

Asked by Sakshi | 1 year ago |  38

##### Solution :-

Given:

The points (1, 5, 7) and (5, 1, -4)

We know x = 0 and y = 0 on z-axis

Let R(0, 0, z) any point on z-axis

According to the question:

RA = RB

RA2 = RB2

By using the formula,

The distance between any two points (a, b, c) and (m, n, o) is given by,

We know, RA2 = RB2

26+ (z – 7)2 = (z + 4)2 + 26

z2+ 49 – 14z + 26

= z2+ 16 + 8z + 26

49 – 14z = 16 + 8z

49 – 16 = 14z + 8z

22z = 33

z =$$\dfrac{33}{22}$$

$$\dfrac{3}{2}$$

The point R (0, 0,$$\dfrac{3}{2}$$) on z-axis is equidistant from (1, 5, 7) and (5, 1, -4).

Answered by Aaryan | 1 year ago

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