Question

root 2 times the distance between (0,5) and (-5,0) is

Answer 1

$\textbf{Formula used:}$

$\text{The distance between two points }$

$(x_1,y_1)\:\text{and}\:(x_2,y_2)\;\text{is}$

$\boxed{\bf\;d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}$

$\text{Given points are (0,5) and (-5,0)}$

$\text{The distance between (0,5) and (-5,0)}$

$=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$=\sqrt{(0+5)^2+(5-0)^2}$

$=\sqrt{25+25}$

$=\sqrt{25{\times}2}$

$=5\sqrt{2}$

$\text{ \sqrt{2}{\times} distance between (0,5) and (-5,0)}$

$=\sqrt{2}{\times}5\sqrt{2}$

$=2{\times}5$

$=10$

$\therefore\textbf{ \sqrt{2}{\times} distance between (0,5) and (-5,0) is 10}$

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Answer 2

$\sqrt{2}$ times of the distance between (0,5) and (-5,0) is 10 units.

Step-by-step explanation:

The distance formula between two points is given by :

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Here, the points are (0,5) and (-5,0)

Distance between these two points is given by :

$d=\sqrt{((-5)-0)^2+(0-5)^2}\\\\d=\sqrt{25+25} \\\\d=\sqrt{50} \\\\d=5\sqrt{2}$

It is required to find $\sqrt{2}$ times of the distance between these two points. So,

$d=5\sqrt{2} \times \sqrt{2} \\\\d=5\times 2\\\\d=10$

So, $\sqrt{2}$ times of the distance between (0,5) and (-5,0) is 10 units.

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$\sqrt{2}$ times of the distance between (0,5) and (-5,0) is 10 units.

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