Given:
The pair of points.
5. P(1, 1), Q(–1, –1)
6. [tex]E\left(\dfrac{1}{2},4\dfrac{1}{4}\right), F\left(5,-\dfrac{1}{2}\right)[/tex]
To find:
The distance between the pair of points.
Solution:
Distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
5.
The distance between the pair of points P(1, 1) and Q(–1, –1) is:
[tex]PQ=\sqrt{(-1-1)^2+(-1-1)^2}[/tex]
[tex]PQ=\sqrt{(-2)^2+(-2)^2}[/tex]
[tex]PQ=\sqrt{4+4}[/tex]
[tex]PQ=\sqrt{8}[/tex]
[tex]PQ=2\sqrt{2}[/tex]
Therefore, the distance between P and Q is [tex]2\sqrt{2}[/tex].
6.
The distance between the pair of point [tex]E\left(\dfrac{1}{2},4\dfrac{1}{4}\right), F\left(5,-\dfrac{1}{2}\right)[/tex] is:
[tex]EF=\sqrt{\left(5-\dfrac{1}{2}\right)^2+\left(-\dfrac{1}{2}-4\dfrac{1}{4}\right)^2}[/tex]
[tex]EF=\sqrt{\left(\dfrac{10-1}{2}\right)^2+\left(-\dfrac{1}{2}-\dfrac{17}{4}\right)^2}[/tex]
[tex]EF=\sqrt{\left(\dfrac{9}{2}\right)^2+\left(\dfrac{-2-17}{4}\right)^2}[/tex]
[tex]EF=\sqrt{\dfrac{81}{4}+\left(\dfrac{-19}{4}\right)^2}[/tex]
On further simplification, we get
[tex]EF=\sqrt{\dfrac{81}{4}+\dfrac{361}{16}}[/tex]
[tex]EF=\sqrt{\dfrac{324+361}{16}}[/tex]
[tex]EF=\sqrt{\dfrac{685}{16}}[/tex]
[tex]EF=\sqrt{42.8125
}[/tex]
[tex]EF\approx 6.5[/tex]
Therefore, the distance between E and F is 6.5 units.
