Respuesta :

r3t40

The distance is,

[tex]d(A,B)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

Where [tex]A(x_1,y_1),B(x_2,y_2)\longrightarrow A(-3,-2),B(-1,-2)[/tex]

[tex]d(A,B)=\sqrt{(-1-(-3))^2+(-2-(-2))^2}=\sqrt{4}=2[/tex]

The distance between points A, B is 2 units.

Hope this helps.

r3t40

The formula for distance between two points is:

[tex]\sqrt{(x_{2} -x_{1})^{2} + (y_{2} -y_{1})^{2}}[/tex]

In this case:

[tex]x_{2} =-1\\x_{1} =-3\\y_{2} =-2\\y_{1} =-2[/tex]

^^^Plug these numbers into the formula for distance like so...

[tex]\sqrt{(-1 - (-3))^{2} + (-2 - (-2))^{2}}[/tex]

To solve this you must use the rules of PEMDAS (Parentheses, Exponent, Multiplication, Division, Addition, Subtraction)

First we have parentheses. Remember that when solving you must go from left to right

[tex]\sqrt{(-1 - (-3))^{2} + (-2 - (-2))^{2}}[/tex]

-1 - (-3) = 2

[tex]\sqrt{(2)^{2} + (-2 - (-2))^{2}}[/tex]

-2 - (-2) = 0

[tex]\sqrt{(2)^{2} + (0)^{2}}[/tex]

Next solve the exponent. Again, you must do this from left to right

[tex]\sqrt{(2)^{2} + (0)^{2}}[/tex]

2² = 4

[tex]\sqrt{4 + (0)^{2}}[/tex]

0² = 0

[tex]\sqrt{(4+0)}[/tex]

Now for the addition

[tex]\sqrt{(4 + 0)}[/tex]

4 + 0 = 4

√4 <<<This can be further simplified to...

2

***Remember that the above answers are in terms of units

Hope this helped!

~Just a girl in love with Shawn Mendes

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