Big Ideas Math Geometry, 2014
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Big Ideas Math Geometry, 2014 View details
3. Using Midpoint and Distance Formulas
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Exercise 4 Page 19

Practice makes perfect
a We can find the midpoint and length separately.

Midpoint

To find the midpoint, we start by determining the distance between the points in the x and y-directions. Counting the number of units in both directions, we get 24 and 10.

This means that the midpoint of the horizontal line segment is 12 units to the right of D(-10,-4), half of the total x-distance. Similarly, the midpoint of the vertical line segment is 5 units below E(14,6).

This gives us our midpoint at M(2,1).

Length

We can find the length of the segment using the Pythagorean Theorem. DE makes up the hypotenuse of a right triangle with legs that are 24 and 10 units long.
a^2+b^2=c^2
24^2+10^2=DE^2
Solve for DE
576+100=DE^2
676=DE^2
DE^2=676
DE=±sqrt(676)

DE > 0

DE=sqrt(676)
DE=26
The length is 26 units.
b Let's find the midpoint and length separately.

Midpoint

We can begin by marking the points on a coordinate plane and counting the number of units in the x and y-direction.

The lengths of the vertical and horizontal line segments are 13 and 8 units. The midpoint of each segment will be 6.5 and 4 units from the endpoints.

This gives us the midpoint M(2.5,4).

Length

The line segment FG makes up the hypotenuse in a right triangle with legs 8 and 13 units long. Therefore, we use the Pythagorean Theorem to find the length.
a^2+b^2=c^2
13^2+8^2=FG^2
Solve for FG
169+64=FG^2
233=FG^2
FG^2=233
FG=±sqrt(233)

FG > 0

FG=sqrt(233)
FG=15.26433...
FG≈ 15.3
The length of FG is about 15.3 units.