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Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, -4)
CBSE | Class 11 | Exercise 28.2 | Introduction to Three Dimensional Coordinate Geometry | Maths | RD Sharma
Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, -4)

Given:

The points

    \[\left( 1,\text{ }5,\text{ }7 \right)\text{ }and\text{ }\left( 5,\text{ }1,\text{ }-4 \right)\]

We know

    \[x\text{ }=\text{ }0\text{ }and\text{ }y\text{ }=\text{ }0\text{ }on\text{ }z-axis\]

Let

    \[R\left( 0,\text{ }0,\text{ }z \right)\]

any point on z-axis

According to the question:

    \[RA\text{ }=\text{ }RB\]

    \[R{{A}^{2}}~=\text{ }R{{B}^{2}}\]

By using the formula,

The distance between any two points

    \[\left( a,\text{ }b,\text{ }c \right)\text{ }and\text{ }\left( m,\text{ }n,\text{ }o \right)\]

is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 21

We know,

    \[R{{A}^{2}}~=\text{ }R{{B}^{2}}\]

    \[26+\text{ }{{\left( z\text{ }\text{ }7 \right)}^{2}}~=\text{ }{{\left( z\text{ }+\text{ }4 \right)}^{2}}~+\text{ }26\]

Or,

    \[{{z}^{2}}+\text{ }49\text{ }-\text{ }14z\text{ }+\text{ }26\text{ }=\text{ }{{z}^{2}}+\text{ }16\text{ }+\text{ }8z\text{ }+\text{ }26\]

    \[49\text{ }-\text{ }14z\text{ }=\text{ }16\text{ }+\text{ }8z\]

Or,

    \[49\text{ }-\text{ }16\text{ }=\text{ }14z\text{ }+\text{ }8z\]

    \[22z\text{ }=\text{ }33\]

Or,

    \[z\text{ }=\text{ }33/22\]

    \[=\text{ }3/2\]

∴The point

    \[R\text{ }\left( 0,\text{ }0,\text{ }3/2 \right)\]

on z-axis is equidistant from 

    \[\left( 1,\text{ }5,\text{ }7 \right)\text{ }and\text{ }\left( 5,\text{ }1,\text{ }-4 \right)\]

i

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