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2 years ago

What is the distance between (7, -7) and (2, 5) on the coordinate plane? 5 units 17 units 7 units 13 units

Mathematics

1 answer:

olchik [2.2K]2 years ago

3 0

The answer is, "13 units."

You can calculate this by using the distance formula.

d=√(x₂-x₁)²+(y₂-y₁)²

d=√(2-7)²+(5-(-7))²

d=√-5²+12²

d=√25+144

d=√169

d=13

The distance between (7, -7) and (2, 5) is 13 units.

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$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ R(\stackrel{x_1}{-7}~,~\stackrel{y_1}{5})\qquad S(\stackrel{x_2}{17}~,~\stackrel{y_2}{5}) ~\hfill RS=\sqrt{(~~ 17- (-7)~~)^2 + (~~ 5- 5~~)^2} \\\\\\ ~\hfill RS=\sqrt{( 24)^2 + ( 0)^2}\implies \boxed{RS=24} \\\\\\ S(\stackrel{x_1}{17}~,~\stackrel{y_1}{5})\qquad T(\stackrel{x_2}{5}~,~\stackrel{y_2}{0}) ~\hfill ST=\sqrt{(~~ 5- 17~~)^2 + (~~ 0- 5~~)^2}$

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