What is the distance between the points (2,-3) and (-6,4) on the coordinate plane?

[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{-6}~,~\stackrel{y_2}{4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[-6-2]^2+[4-(-3)]^2}\implies d=\sqrt{(-6-2)^2+(4+3)^2} \\\\\\ d=\sqrt{64+49}\implies d=\sqrt{113}\implies d\approx 10.63[/tex]

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sandlee09 sandlee09 · Answer 2 · 2019-07-22T20:18:37+02:00

Answer:

The distance between both points on the coordinate plane is 10.63

Step-by-step explanation:

We can visualize the problem in form of a triangle to better understand the situation. As soon in the picture below, we need to find the hypotenuse (h) which is the distance between both points. In order to do this we first need to find the value of both x and y.

[tex]x = 2 - (-6)[/tex]

[tex]x = 2+6[/tex]

[tex]x = 8[/tex]

[tex]y = -3 - 4[/tex]

[tex]y = -7[/tex]

Now that we have boy the length of x and y , we can use the Pythagorean Theorem to solve for the hypotenuse.

[tex]x^{2} +y^{2} =h^{2}[/tex]

[tex]8^{2} +(-7)^{2} =h^{2}[/tex]

[tex]64+47 =h^{2}[/tex]

[tex]113=h^{2}[/tex]

[tex]10.63 = h[/tex]

Finally, we can see that the distance between both points on the coordinate plane is 10.63

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