Respuesta :

Answer:

[tex]d = \sqrt[2]{34}[/tex]

Step-by-step explanation:

We are solving for the segment of AB. Note that it is a line segment, so there will be end points, those being A(-4, 5) & B(2, -5).

Use the following distance formula:

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

Let:

Point B(2 , -5) = (x₁ , y₁)

Point A(-4 , 5) = (x₂ , y₂)

Plug in the corresponding numbers to the corresponding variables.

[tex]d =[/tex] [tex]\sqrt{(-4 - 2)^{2} + (5 - (-5))^{2} }[/tex]

Simplify. Remember to follow PEMDAS. First, solve the parenthesis, then the powers, then add, and then finally square root.

[tex]d = \sqrt{(-6)^{2} + (5 + 5)^{2} } \\d = \sqrt{(-6 * -6) + (10)^2} \\d = \sqrt{(36) + (10 * 10)} \\d = \sqrt{36 + (100)} \\d = \sqrt{136}[/tex]

Simplify:

[tex]d = \sqrt{136} = \sqrt{2 * 2 * 2 * 17} = \sqrt[2]{17 * 2} = \sqrt[2]{34}[/tex]

[tex]d = \sqrt[2]{34}[/tex] is your answer.

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