Question

Find the distance between the points (-3, -2) and (-1, -2).

Answer 1

The formula for distance between two points is:

$\sqrt{(x_{2} -x_{1})^{2} + (y_{2} -y_{1})^{2}}$

In this case:

$x_{2} =-1\\x_{1} =-3\\y_{2} =-2\\y_{1} =-2$

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

$\sqrt{(-1 - (-3))^{2} + (-2 - (-2))^{2}}$

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

$\sqrt{(-1 - (-3))^{2} + (-2 - (-2))^{2}}$

-1 - (-3) = 2

$\sqrt{(2)^{2} + (-2 - (-2))^{2}}$

-2 - (-2) = 0

$\sqrt{(2)^{2} + (0)^{2}}$

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

$\sqrt{(2)^{2} + (0)^{2}}$

2² = 4

$\sqrt{4 + (0)^{2}}$

0² = 0

$\sqrt{(4+0)}$

Now for the addition

$\sqrt{(4 + 0)}$

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

Answer 2

The distance is,

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

Where $A(x_1,y_1),B(x_2,y_2)\longrightarrow A(-3,-2),B(-1,-2)$

$d(A,B)=\sqrt{(-1-(-3))^2+(-2-(-2))^2}=\sqrt{4}=2$

The distance between points A, B is 2 units.

Hope this helps.

r3t40