Using the Distance and Midpoint Formulas in Proofs

We want to find the midpoint of the line segment with the given endpoints. For this purpose, we can use the Midpoint Formula.

The midpoint is at (8,8).

Like in Part A, we want to find the midpoint of the line segment with the given endpoints. To do so, we will substitute the given points into the Midpoint Formula and evaluate.

The midpoint of the line segment is (8,6)

As in previous parts, we will substitute the given points into the Midpoint Formula.

As we can see, the midpoint is (1,9).

Like in previous parts, we should substitute the given endpoints into the Midpoint Formula to determine the midpoint.

The line segment has its midpoint at (2, 4).

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Segment	Points	Substitute	Simplify
$\overline{A C}$	$A (- 3, 0)$ $&$ $C (4, - 1)$	$A C =$ $(4 - (- 3))^{2} + (- 1 - 0)^{2}$	$A C =$ $52$
$\overline{A B}$	$A (- 3, 0)$ $&$ $B (- 2, 2)$	$A B =$ $(- 2 - (- 3))^{2} + (2 - 0)^{2}$	$A B =$ $5$
$\overline{B C}$	$B (- 2, 2)$ $&$ $C (4, - 1)$	$B C =$ $(4 - (- 2))^{2} + (- 1 - 2)^{2}$	$B C =$ $35$

$M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Diagonal	Endpoints	Substitute	Simplify
$\overline{B D}$	$B (- 2, 2)$ $&$ $D (3, - 3)$	$M (\frac{- 2 + 3}{2}, \frac{2 + ( - 3 )}{2})$	$M (0.5, - 0.5)$ $✓$
$\overline{A C}$	$A (- 3, 0)$ $&$ $D (4, - 1)$	$M (\frac{- 3 + 4}{2}, \frac{0 + ( - 1 )}{2})$	$M (0.5, - 0.5)$ $✓$

$M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Side	Endpoints	Substitute	Simplify
$\overline{A B}$	$A (0, 0)$ and $B (2, 2.5)$	$M_{1} (\frac{0 + 2}{2}, \frac{0 + 2 . 5}{2})$	$M_{1} (1, 1.25)$
$\overline{B C}$	$B (2, 2.5)$ and $C (6, 0)$	$M_{2} (\frac{2 + 6}{2}, \frac{2 . 5 + 0}{2})$	$M_{2} (4, 1.25)$

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Segment	Endpoints	Substitute	Simplify
$\overline{M_{1} M_{2}}$	$M_{1} (1, 1.25)$ and $M_{2} (4, 1.25)$	$M_{1} M_{2} = (4 - 1)^{2} + (1.25 - 1.25)^{2}$	$M_{1} M_{2} = 3$
$\overline{B C}$	$A (0, 0)$ and $C (6, 0)$	$A C = (6 - 0)^{2} + (0 - 0)^{2}$	$A C = 6$

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Segment	Points	Substitute	Simplify
$\overline{A D}$	$A (0, 0)$ $&$ $D (3, 0)$	$A D =$ $(3 - 0)^{2} + (0 - 0)^{2}$	$A D =$ $3$
$\overline{B C}$	$B (2, 2)$ $&$ $C (5, 2)$	$B C =$ $(5 - 2)^{2} + (2 - 2)^{2}$	$B C =$ $3$

Using the Distance and Midpoint Formulas in Proofs

Catch-Up and Review

Investigating the Properties of a Parallelogram

Distance Formula

Proof

Practice Finding the Distance Between Two Coordinate Pairs

Classifying a Parallelogram Using Its Coordinates

Hint

Solution

Determining If a Point Is On a Circle's Circumference

Hint

Solution

Definition of a Midpoint

Midpoint

Midpoint Formula

Proof

Practice Finding Midpoints Using the Midpoint Formula

Proving the Properties of a Parallelogram

Answer

Hint

Solution

Using Rigid Motions to Prove a Theorem

Extra

Proving the Triangle Midsegment Theorem by Rigid Motions

Answer

Hint

Solution

Proving the Properties of a Parallelogram

Answer

Hint

Solution

Slopes

Lengths

Using the Distance and Midpoint Formulas in Proofs

Recommended exercises

	12 Theory slides
	15 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Segment	Points	y_2-y_1/x_2-x_1	m
MN	N(- 2,2), M(1,7)	7- 2/1-( - 2)	5/3
NP	N(- 2,2), P(3,- 1)	- 1- 2/3-( - 2)	- 3/5

Segment	Points	y_2-y_1/x_2-x_1	m
MQ	M(1,7), Q(6,4)	4- 7/6- 1	- 3/5
QP	Q(6,4), P(3,- 1)	- 1- 4/3- 6	5/3

Segment	Points	sqrt((x_2-x_1)^2+(y_2-y_1)^2)	d
MN	M(1,7), N(- 2,2)	sqrt(( 1-( - 2))^2+( 7- 2)^2)	sqrt(34)
QP	Q(6,4), P(3,- 1)	sqrt(( 6- 3)^2+( 4-( - 1))^2)	sqrt(34)
NP	N(- 2,2), P(3,- 1)	sqrt(( - 2- 3)^2+( 2-( - 1))^2)	sqrt(34)
MQ	M(1,7), Q(6,4)	sqrt(( 1- 6)^2+( 7- 4)^2)	sqrt(34)

Triangle Midsegment Theorem
The segment that connects the midpoints of two sides of a triangle — a midsegment — is parallel to the third side of the triangle and half its length.