Find the distance between the two points. (2, 5) and (–1, –5)

2 answers:

6 0

√109≈10.44030650 <--- Answer

Distance=√(x2−x1)2+(y2−y1)2

√((−1)−(2))2+((−5)−(5))2

That check sign is the square root, since they wouldnt let me put it in...

mario62 [17]3 years ago

4 0

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 2 &,& 5~) 
% (c,d)
&&(~ -1 &,& -5~)
\end{array}
\\\\\\
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
d=\sqrt{(-1-2)^2+(-5-5)^2}
\\\\\\
d=\sqrt{(-3)^2+(-10)^2}\implies d=\sqrt{109}$

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The end points of one side of a regular pentagon are (-1,4) and (2,3). What is the perimeter of the pentagon?

marissa [1.9K]

To determine the perimeter of the pentagon, you must first calculate a side length of it. Let's name the coordinates A(-1,4) and B(2,3).

To figure out how far the points are from each other, you have to use the distance formula:
$D_{AB} = \sqrt{( x_{2}-x_{1})^2+{(y_{2}-y_{1})^2}
$
$x_{1} =1, x_{2} =-2, y_{1} =2, y_{2} =3$
$D_{AB}= \sqrt{(2--1)^2+{(3-4)^2}$
D_{AB}= \sqrt{(2--1)^2+{(3-4)^2}
$D_{AB}= \sqrt{(2+1)^2+{(3-4)^2}$
$D_{AB}= \sqrt{(3)^2+(-1)^2}$
$D_{AB}= \sqrt{9+1}$
$D_{AB}= \sqrt{10}$

Now, the formula for the perimeter of a pentagon is
P = 5×side length

So...
Perimeter = 5× $\sqrt{10}$

The answer is (2)

8 0

3 years ago

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slavikrds [6]

Answer:

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