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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