Let P (x_1,y_1,z_1) and Q (x_2,y_2,z_2) be points in 3 space. Let M (\frac{x_1+x_2}{2},...


Let {eq}P (x_1,y_1,z_1){/eq} and {eq}Q (x_2,y_2,z_2){/eq} be points in {eq}3{/eq} space. Let {eq}M (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}){/eq}. Verify {eq}PM = QM {/eq}

Distance Between Points

Let A(x_1,x_2,x_3) and B (Y_!,Y_2,Y_3) be two points in 3D space then distance between these point can be calculated from this formula.

{eq}AB=\sqrt{(Y_1-X_1)^2+(Y_2-X_2)^2+(Y_3-X_3)^2} \\ {/eq}

If one point is from same distance from another points its called equidistant point .

Answer and Explanation:

We have to prove P (x_1,y_1,z_1) and Q (x_2,y_2,z_2) are equidistant from point M.

Let first find PM

{eq}PM=\sqrt{(\displaystyle\frac{x_1+x_2}{2}-x_1)^2+( \displaystyle\frac{y_1+y_2}{2}-y_1)^2+( \displaystyle\frac{z_1+z_2}{2})-z_1)^2} \\ PM=\sqrt{(\displaystyle\frac{x_2-x_1}{2})^2+(\displaystyle\frac{y_2-y_1}{2})^2+(\displaystyle\frac{z_2-z_1}{2})^2} \\ {/eq}

Now calculating QM

{eq}QM=\sqrt{(\displaystyle\frac{x_1+x_2}{2}-x_2)^2+( \displaystyle\frac{y_1+y_2}{2}-y_2)^2+( \displaystyle\frac{z_1+z_2}{2})-z_2)^2} \\ QM=\sqrt{(\displaystyle\frac{x_1-x_2}{2})^2+(\displaystyle\frac{y_1-y_2}{2})^2+(\displaystyle\frac{z_1-z_2}{2})^2} \\ QM=\sqrt{(\displaystyle\frac{-(x_2-x_1)}{2})^2+(\displaystyle\frac{-(y_2-y_1)}{2})^2+(\displaystyle\frac{-(z_2-z_1)}{2})^2} \\ {/eq}

Clearly PM=QM

Learn more about this topic:

How to Find the Distance Between a Point & a Line

from FTCE Mathematics 6-12 (026): Practice & Study Guide

Chapter 30 / Lesson 5

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