To recall, real numbers consist of integers, fractions, rational and irrational numbers which we can represent on the real number line. When we move a positive direction on the number line it is almost always moving up or two the right, when we move in a negative direction it is almost always to the left.

On the number line, if we move from x = a to x = b then distance moved is b - a. This change of *x* is symbolized by Δx.

This symbol, (delta) Δ is what we use to symbolize a change in something. So,

For moving from x = a to x = b, the change in *x*, *a* to *b* is represented like,

Δx = b - a

If x = b is a bigger value than x = a, then Δx = b - a will be positive, meaning that the move from a to b is in a positive direction.

If x = b is a smaller value than x = a, then Δx = b - a will be negative, meaning that the move from a to b is in a negative direction.

If x = b is the same as x = a, then Δx = b - a will be zero, meaning that there was no movement of the points.

We also use Δ when representing the distance between two points a and b, for this we take the absolute value of Δ, or |Δ|, meaning strictly the positive value of the number. ( |-3| = 3 )

So:

DEFINITION

dist(a, b) = b - a if b ≥ a or a - b if b > a

Or, using our absolute value (the positive value),

dist(a, b) = |Δx| = |b - a| =

(Δ x)^{2}

DEFINITION

The Midpoint is the point halfway between two points, we calculate this by,

M =

a + b

2

Question: Find the distance and the midpoint of the points x = -1 and x = 7,

Recall our definition, the distance between two points, a and b, on a number line is,

dist(a, b) = |Δx| = |b - a| =

(Δ x)^{2}

We want to know how far away x = -1 is from x = 7,

[if we just calculated Δ x then that would give us the direction (if it was positive or negative) as well as the distance, which we don't want.]

For this we want the absolute value of Δ x, that is, the positive value of Δ x.

So,

dist(-1, 7) = |delta x| = |7 - (-1)| = |8| = 8

So the distance is 8

For the Midpoint,

M =

a + b

2

M =

-1 + 7

2

M =

6

2

M = 3

So the midpoint between x = -1 and x = 7 is x = 3

A Cartesian plane is defined by two perpendicular number lines: the x-axis, which is horizontal, and the y-axis, which is vertical. Using these axes, we can describe any point in the plane using an ordered pair of numbers, usually *x* and *y*, said like (x, y). A Cartesian plane is thought of as two dimensional.

If we have to points, *A* at (x_{1}, y_{1}) and *B* at (x_{2}, y_{2}) on a Cartesian plane and we want to calculate the distance between them, we approach it the same way we would when calculating two points on a number line, except we have to account for the fact we have two dimensions rather than one.

The horizontal direction of the change in x-values is:

Δ x = x_{2} - x_{1}

The vertical direction of the change in y-values is

Δ y = y_{2} - y_{1}

For our distance between *A* and *B* recall the Pythagorean theorem of,

a^{2} + b^{2} = c ^{2},

In this case, the c value from the theorem is equivalent to dist(A, B), and the Δ x and Δ y are equivalent to the *a* and *b* value from the theorem,

(dist(A, B))^{2} = (Δ x)^{2} + (Δ y)^{2}

DEFINITION

dist(A, B) =

(Δ x)^{2} + (Δ y)^{2}

DEFINITION

The Midpoint, *M* of *A* and *B* is:

M =

x_{1} + x_{2}

2

,

y_{1} + y_{2}

2

Question: Calculate the distance and midpoint between points *T* (2, 3) and *R* (8, 10) to the nearest tenth of a decimal place

So, our equation for distance between two points on two dimensional Cartesian plane is,

dist(A, B) =

(Δ x)^{2} + (Δ y)^{2}

So for,

dist(T, R) =

(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

dist(T, R) =

(8 - 2)^{2} + (10 - 3)^{2}

dist(T, R) =

6^{2} + 7^{2}

dist(T, R) =

36 + 49

dist(T, R) =

85

= 9.2

The midpoint is,

M =

x_{1} + x_{2}

2

,

y_{1} + y_{2}

2

M =

2 + 8

2

,

3 + 10

2

M =

10

2

,

13

2

(5, 6.5)

Question: Find an equation for the line with all the points of P = (x, y) that are an equal distance from S = (1, 2) and T = (4, -2).

For this question, we are looking for all the points that are a equal distance from the points *S* and *T*, these points will all be on a straight line,

Note, that for every point on the line with all the equal distant points, each point on that line is the same distance from *S* as it is from *T* (d1, d2, d3 etc…).

Any point that is not on that line will be closer to either the point *T* or *S*.

Now, lets do the math,

We know that the general formula for distance between two points *A*(x_{1}, y_{1}) and *B*(x_{2}, y_{2}) is,

dist(A, B) =

(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

And we know that all the points *P*(x, y) are the same distance from *S* as they are from *T*,

The distance from *P* to *T* is equal to the distance from *P* to *S*, so,

dist(P, T) = dist(P, S)

Now, we sub in the values of P(x, y) - which will give us a general equation for the straight line between the two points *T* and *S*. We also sub in our T(4, -2) and S(1, 2) into,

dist(P, T) = dist(P, S)

dist(P, T) =

(x - 4)^{2} + (y - (-2)^{2}

dist(P, S) =

(x - 1)^{2} + (y - 2)^{2}

Now equating those two equations,

(x - 4)^{2} + (y - (-2)^{2}

=

(x - 1)^{2} + (y - 2)^{2}

Now, squaring both sides of the equation we get,

(x - 4)^{2} + (y + 2)^{2} = (x - 1)^{2} + (y - 2)^{2}

x^{2} -8x + 16 + y^{2} + 4y + 4 = x^{2} - 2x + 1 + y^{2} - 4y + 4

Subtract x^{2} and y^{2} from both sides of the equation,

-8x + 16 + 4y + 4 = - 2x + 1 - 4y + 4

-8x + 20 + 4y = -2x + 5 - 4y

Add 2y to both sides,

-8x + 20 + 8y = -2x + 5

Add 8x and subtract 20,

8y = 6x - 15

Divide both sides by 8,

y =

6x - 15

8

There we have our equation for the line that runs between the points *S* and *T*, every point on that line will be an equal distance from *S* and *T*.