Question

Use the Distance Formula to find the distance between each pair of points. Round to the nearest tenth if necessary.

Answer 1

Given:

The pair of points.

5. P(1, 1), Q(–1, –1)

6. $E\left(\dfrac{1}{2},4\dfrac{1}{4}\right), F\left(5,-\dfrac{1}{2}\right)$

To find:

The distance between the pair of points.

Solution:

Distance formula:

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

5.

The distance between the pair of points P(1, 1) and Q(–1, –1) is:

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

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

$PQ=\sqrt{4+4}$

$PQ=\sqrt{8}$

$PQ=2\sqrt{2}$

Therefore, the distance between P and Q is $2\sqrt{2}$.

6.

The distance between the pair of point $E\left(\dfrac{1}{2},4\dfrac{1}{4}\right), F\left(5,-\dfrac{1}{2}\right)$ is:

$EF=\sqrt{\left(5-\dfrac{1}{2}\right)^2+\left(-\dfrac{1}{2}-4\dfrac{1}{4}\right)^2}$

$EF=\sqrt{\left(\dfrac{10-1}{2}\right)^2+\left(-\dfrac{1}{2}-\dfrac{17}{4}\right)^2}$

$EF=\sqrt{\left(\dfrac{9}{2}\right)^2+\left(\dfrac{-2-17}{4}\right)^2}$

$EF=\sqrt{\dfrac{81}{4}+\left(\dfrac{-19}{4}\right)^2}$

On further simplification, we get

$EF=\sqrt{\dfrac{81}{4}+\dfrac{361}{16}}$

$EF=\sqrt{\dfrac{324+361}{16}}$

$EF=\sqrt{\dfrac{685}{16}}$

$EF=\sqrt{42.8125 }$

$EF\approx 6.5$

Therefore, the distance between E and F is 6.5 units.