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Find the distances from (-2,-4) to each of (-6, 8), (5,2) and (9, 5).

Guest May 19, 2017
 #1
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First, know the distance formula:

 

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

 

Now, simply substitute the values into this formula and simplify:

 

\(d=\sqrt{(-6-(-2))^2+(8-(-4))^2}\)

\(=\sqrt{(-6+2)^2+(8+4)^2}\)

\(=\sqrt{(-4)^2+12^2}\)

\(=\sqrt{16+144}\)

\(=\sqrt{160}\)

 

Of course, put all radicals in simplest radical form. I also included an approximate decimal to the ten thousandths place:

 

\(\sqrt{160}=\sqrt{16*10}=\sqrt{16}\sqrt{10}=4\sqrt{10}\approx12.6491\)

 

Let's continue this process for the rest of the points. I'll do the distance from (-2,-4) to (-6,8):

 

\(d=\sqrt{(-5-(-2))^2+(2-(-4))^2}\)

\(=\sqrt{(5+2)^2+(2+4)^2}\)

\(=\sqrt{7^2+6^2}\)

\(=\sqrt{49+36}\)

\(=\sqrt{85}\approx9.220\)

 

The square root of 85 is already in simplest radical form, so no need to do anymore.

 

 


 

Guest May 19, 2017
 #2
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The previous didn't find the final distance from (-2,-4) to (9,5), so I will:

 

\(d=\sqrt{(9-(-2))^2+(-5-(-4))^2}\)

\(=\sqrt{(9+2)^2+(-5+4)^2}\)

\(=\sqrt{11^2+(-1)^2}\)

\(=\sqrt{121+1}\)

\(=\sqrt{122}\approx11.0454\)

 

The square root of 122 has no perfect square factors, so it is already in simplest radical form.

Guest May 19, 2017

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