Distance Formula and Midpoint Formula: Algebraic Proofs in Geometry

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

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})

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)

✓

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.

$M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
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 $C (6, 0)$	$M_{2} (\frac{2 + 6}{2}, \frac{2 . 5 + 0}{2})$	$M_{2} (4, 1.25)$

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

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
$\overline{A D}$	$A (0, 0)$ $&$ $D (3, 0)$	$A D =$ $(3 - 0)^{2} + (0 - 0)^{2}$	$A D =$ $3$
$\overline{B C}$	$B (2, 2)$ $&$ $C (5, 2)$	$B C =$ $(5 - 2)^{2} + (2 - 2)^{2}$	$B C =$ $3$

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

Below we see a circle with its center at the origin. On the circle, an arbitrary point $(x, y)$ has been plotted.

A circle with a radius of 4 and a point (x,y) plotted on it

What should

x^{2} + y^{2}

equal? Find your answer without using the equation of a circle.

Examining the diagram, we notice that the circle has a radius r= 4 units. This means the distance d between the origin and every point on the circle, including the plotted point, is 4 units.

We can write an equation by using the Distance Formula for the points (0,0) and (x,y).

As we can see, x^2+y^2 equals 16.

A B C D

is a parallelogram. If

A (3, - 4),

B (6, 2),

and

C (4, 6),

then what are the possible locations of point

D ?

A parallelogram is a quadrilateral with two pairs of parallel sides. In ABCD, we have been given three vertices. Let's place these in a coordinate plane.

We have three possibilities here.

Let's go through these one at a time.

1) AC and BC are Sides

When AC and BC are sides, the last point must be placed such that AC∥ DB and BC∥ DA. To find this point, we can measure the vertical and horizontal distance between C and B. By moving the same vertical and horizontal distance from point A, we can place the fourth point D.

2) AB and BC are Sides

When AB and BC are sides, the last point must be placed such that AB∥ DC and BC∥ AD. We will measure the vertical and horizontal distance between B and C. By moving the same vertical and horizontal distance from point A, we can place the fourth point, D.

3) AC and AB are Sides

This time, D must be placed such that AC∥ BD and AB∥ CD. We will measure the vertical and horizontal distance between A and B. By moving the same vertical and horizontal distance from point C, we can place point D.

Conclusion

The possible positions of the fourth vertex D are (5,- 8), (1,0), and (7,12).

Consider $△ A B C .$

What would you call this triangle based on its angles?

What would you call this triangle based on its sides?

Determine

m ∠ A .

Let's begin by calculating the slope of each segment using the Slope Formula.

Segment	Points	y_2-y_1/x_2-x_1	m
AB	A(3,0), B(2,7)	7- 0/2- 3	-7
AC	A(3,0), C(6,4)	4- 0/6- 3	4/3
BC	B(2,7), C(6,4)	4- 7/6- 2	- 3/4

We can see that AC and BC are negative reciprocals, which means their slopes multiply to - 1. 4/3(- 3/4)=- 1 When two segments are negative reciprocals they are perpendicular. Therefore, this is a right triangle.

To label the triangle based on its sides, we will first use the Distance Formula to determine the length of each side.

Segment	Points	sqrt((x_2-x_1)^2+(y_2-y_1)^2)	d
AB	A(3,0), B(2,7)	sqrt(( 3- 2)^2+( 0- 7)^2)	5sqrt(2)
AC	A(3,0), C(6,4)	sqrt(( 3- 6)^2+( 0- 4)^2)	5
BC	B(2,7), C(6,4)	sqrt(( 2- 6)^2+( 7- 4)^2)	5

Since BC and AC have the same length, this is an isosceles triangle.

From previous parts, we have figured out that this is a right isosceles triangle. Therefore, according to the Base Angles Theorem, it has two congruent base angles. By using the Interior Angles Theorem we can write the following equation. m∠ A+m∠ B+90^(∘)=180^(∘) Since m∠ B=m∠ A, we can solve this equation for m∠ A if we substitute m∠ A for m∠ B.

The measure of ∠ A is 45^(∘).

On the line $y = 2 x$ there is a point $P$ in the first quadrant whose distance to the origin is $24$ units. Calculate the $x -$ coordinate of $P .$ Answer in exact form.

A straight line with a point P marked on it.

We know that the distance from the origin to P is 24 units. We can let a represent the x-coordinate of the point. Since P is on the line y=2x, the corresponding y-coordinate is 2a.

Let's substitute the given points (a, 2a) and (0, 0) into the Distance Formula and simplify.

Notice that we only kept the positive solution. Since we know that P is in the first quadrant, the x-coordinate is positive. Since the distance to P is 24 units, we can substitute 24 for d and solve for a.

The x-coordinate is 24sqrt(5).

Two points are located at $(a, c)$ and $(b, c) .$

What is the midpoint between these points?

What is the distance between these points?

We can find the midpoint by substituting the given points into the Midpoint Formula.

To find the distance between the points, we use the Distance Formula.

The distance between the points is |b-a|.

A parallelogram

A B C D

has the following coordinates.

A (0, 0), B (6, 0), C (8, 6), D (2, 6)

What is the best name for

A B C D ?

Write the equation of the line that coincides with the diagonal

\overline{A C} .

Let's mark the four points in a coordinate plane and connect them. We already know that this forms a parallelogram so we can mark opposite sides as parallel.

To determine the best name for our parallelogram, let's review the classification of different parallelograms.

	Definition
Parallelogram	Both pairs of opposite sides are parallel
Rhombus	Parallelogram with four congruent sides
Rectangle	Parallelogram with four right angles
Square	Parallelogram with four congruent sides and four right angles

Already, we can tell that the parallelogram can neither be a rectangle nor a square. This requires it to have four right angles and we can see that it does not have any.

Rhombus or Parallelogram?

To choose between a rhombus and a parallelogram, we must determine the length of its four sides. By definition, any pair of opposite sides in a parallelogram are congruent. Therefore, if we know the length of two non-opposite sides it will be sufficient. Let's first find the length of AB which is the difference between the endpoint's x-coordinates since is is horizontal.

As we can see, the pair of horizontal sides are 6 units each. Next, we will use the distance formula to find the length of one of the remaining pair of opposite sides.

As we can see AD and BC are both 2sqrt(10) which means this can not be a rhombus. Therefore, we will label it a parallelogram.

Let's draw the line that falls on the diagonal AC.

Notice that the line passes through the origin. Therefore, we can write it on the following form. y=mx To determine the slope, we substitute the two points in the slope-formula and evaluate.

Now we can write the equation of the line. y=3/4x

Using the Distance and Midpoint Formulas in Proofs

Catch-Up and Review

Investigating the Properties of a Parallelogram