Show that the point $ (7,5) $ is equidistant from the points $ (2,4) $ and $ (6,10) $.

To do:

We have to show that the point \( (7,5) \) is equidistant from the points \( (2,4) \) and \( (6,10) \).

Solution:

We know that,

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

Therefore,

The distance between $(7, 5)$ and \( (2, 4) \) \( =\sqrt{(2-7)^{2}+(4-5)^{2}} \)

$=\sqrt{(-5)^2+(-1)^2}$

$=\sqrt{25+1}$

$=\sqrt{26}$

The distance between $(7, 5)$ and \( (6, 10) \) \( =\sqrt{(6-7)^{2}+(10-5)^{2}} \)

$=\sqrt{(-1)^2+(5)^2}$

$=\sqrt{1+25}$

$=\sqrt{26}$

Hence proved.

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Updated on: 10-Oct-2022

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