Question

What is the distance between A(-8, 4) and B(4, -1)?

Answer 1

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:

$\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2$

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:

$d=\sqrt{(4-(-8))^2+(-1-4)^2}$

Evaluate:

$d=\sqrt{(12)^2+(-5)^2}$

So:

$d=\sqrt{144+25}=\sqrt{169}=13\text{ units}$

The distance between A(-8, 4) and B(4, -1) is 13 units.

Answer 2

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Problem:

• What is the distance between A(-8,4) and B(4,-1)

Given:

$\quad\quad\quad\quad\tt{A.) x\tiny{1}\small{=-8}, y\tiny{1}\small{=4}}$

$\quad\quad\quad\quad\tt{B.) x\tiny{2}\small{=4}, y\tiny{2}\small{=-1}}$

Formula for distance (d):

$\quad\quad\quad\quad\tt{d = \sqrt{(x \tiny{2} \small{ - x \tiny{1} \small {)}^{2} + (y \tiny{2} \small{ - y \tiny{1} \small{)}^{2} } }}}$

Solution:

$\quad\quad\quad\quad\tt{d = \sqrt{(4 - \small{ (- 8}{))}^{2} + ( \small{- 1)}\small{ - {4)}}^{2} }}$

$\quad\quad\quad\quad\tt{d = \sqrt{ ( {12)}^{2} + {( -5)}^{2} }}$

$\quad\quad\quad\quad\tt{d = \sqrt{ {144} + {25}}}$

$\quad\quad\quad\quad\tt{d = \sqrt{ 169}}$

$\quad\quad\quad\quad\tt{d = 13}$

So the final answer is:

$\quad\quad\quad\quad\boxed{\boxed{\tt{\color{magenta}d = 13}}}$

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✍︎ C.Rose❀