Segment [tex]$s_1$[/tex] has endpoints at [tex]$(3+\sqrt{2},5)$[/tex] and[tex]$(4,7)$[/tex]. Segment [tex]$s_2$[/tex] has endpoints at [tex]$(6-\sqrt{2},3)$[/tex] and[tex]$(3,5)$[/tex]. Find the midpoint of the segment with endpoints at the midpoints of [tex]$s_1$[/tex] and [tex]$s_2$[/tex]. Express your answer as [tex]$(a,b)$[/tex].

Respuesta :

Answer:

The midpoint of the segment with endpoints at the midpoints of s1 and s2 is (4,5).

Step-by-step explanation:

Midpoint of a segment:

The coordinates of the midpoint of a segment are the mean of the coordinates of the endpoints of the segment.

Midpoint of s1:

Using the endpoints given in the exercise.

[tex]x = \frac{3 + \sqrt{2} + 4}{2} = \frac{7 + \sqrt{2}}{2}[/tex]

[tex]y = \frac{5 + 7}{2} = \frac{12}{2} = 6[/tex]

Thus:

[tex]M_{s1} = (\frac{7 + \sqrt{2}}{2},6)[/tex]

Midpoint of s2:

[tex]x = \frac{6 - \sqrt{2} + 3}{2} = \frac{9 - \sqrt{2}}{2}[/tex]

[tex]y = \frac{3 + 5}{2} = \frac{8}{2} = 4[/tex]

Thus:

[tex]M_{s2} = (\frac{9 - \sqrt{2}}{2}, 4)[/tex]

Find the midpoint of the segment with endpoints at the midpoints of s1 and s2.

Now the midpoint of the segment with endpoints [tex]M_{s1}[/tex] and [tex]M_{s2}[/tex]. So

[tex]x = \frac{\frac{7 + \sqrt{2}}{2} + \frac{9 - \sqrt{2}}{2}}{2} = \frac{16}{4} = 4[/tex]

[tex]y = \frac{6 + 4}{2} = \frac{10}{2} = 5[/tex]

The midpoint of the segment with endpoints at the midpoints of s1 and s2 is (4,5).

ACCESS MORE
EDU ACCESS