Answer:
The distance between A(-8, 4) and B(4, -1) is 13 units.
Step-by-step explanation:
To find the distance between any two points, we can use the distance formula given by:
[tex]\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2[/tex]
We have the two points A(-8, 4) and B(4, -1). Let A(-8, 4) be (x₁, y₁) and let B(4, -1) be (x₂, y₂). Substitute:
[tex]d=\sqrt{(4-(-8))^2+(-1-4)^2}[/tex]
Evaluate:
[tex]d=\sqrt{(12)^2+(-5)^2}[/tex]
So:
[tex]d=\sqrt{144+25}=\sqrt{169}=13\text{ units}[/tex]
The distance between A(-8, 4) and B(4, -1) is 13 units.
___________________________________
[tex]\quad\quad\quad\quad\tt{A.) x\tiny{1}\small{=-8}, y\tiny{1}\small{=4}}[/tex]
[tex]\quad\quad\quad\quad\tt{B.) x\tiny{2}\small{=4}, y\tiny{2}\small{=-1}}[/tex]
[tex]\quad\quad\quad\quad\tt{d = \sqrt{(x \tiny{2} \small{ - x \tiny{1} \small {)}^{2} + (y \tiny{2} \small{ - y \tiny{1} \small{)}^{2} } }}} [/tex]
[tex]\quad\quad\quad\quad\tt{d = \sqrt{(4 - \small{ (- 8}{))}^{2} + ( \small{- 1)}\small{ - {4)}}^{2} }}[/tex]
[tex]\quad\quad\quad\quad\tt{d = \sqrt{ ( {12)}^{2} + {( -5)}^{2} }}[/tex]
[tex]\quad\quad\quad\quad\tt{d = \sqrt{ {144} + {25}}}[/tex]
[tex]\quad\quad\quad\quad\tt{d = \sqrt{ 169}}[/tex]
[tex]\quad\quad\quad\quad\tt{d = 13}[/tex]
[tex]\quad\quad\quad\quad\boxed{\boxed{\tt{\color{magenta}d = 13}}}[/tex]
___________________________________
#CarryOnLearning
✍︎ C.Rose❀
