# Coordinates and Midpoints

## Coordinates and Midpoints Revision

**Coordinates and Midpoints**

The** Cartesian coordinates** are a coordinate system that is in made up of two axes, the x-axis and the y-axis. As such each point on the **x–y plane** has an x-coordinate and a y-coordinate, that places the point in the 2\text{D} space.

**Midpoints** are the halfway points between two points. So if A and C are points on the x–y plane, then we can define a point B exactly half way between them. So that the** distance **between A and B is the same as the distance between B and C.

**Cartesian Coordinates**

For a point A the** Cartesian coordinates **of A can be given as

(A_x, A_y),

where A_x is the** x-coordinate** and A_y is the** y-coordinate.** The x-coordinate always is written first (just remember x comes before y in the alphabet).

We call the point with coordinates (0,0) the **origin**, denoted O.

So if point A has the coordinates (3,2), that means that, from the origin, the point lies 3 units to the right (the x-axis) and 2 units up (the y-axis), as shown.

**Distance Between Points**

One of the many useful things about Cartesian coordinates, is the ease at which we can find the precise** distance** between any two points.

Let point A = (x_1, y_1) and point B = (x_2, y_2). What we can do, between these two points, is construct a **right angled triangle**. The vertices of this triangle will be A, B and the point (x_2, y_1). Then to find the distance from A to B we can find the base of the triangle and the height of the triangle.

The base of this triangle is A to (x_2, y_1). As they have the same y-coordinate, the **base** of this triangle is

\text{Base}= x_2-x_1.

Similarly, the height of the triangle is from (x_2, y_1) to B. As they have the same x-coordinate, the **height** of this triangle is

\text{Height}=y_2-y_1.

Now, we can just use the** Pythagorean theorem**, as we have constructed a right angle triangle,

\text{Distance }AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2},

as the distance between A and B, is just the length of the hypotenuse of the triangle.

For our example here, the points are (2,3) and (6,6), so the base of out triangle is

\text{Base}= 6 - 2 = 4

the height is

\text{Height} =6 - 3= 3

and so by Pythagoras, the distance between the points is

\text{Distance between points }= \sqrt{4^2+3^2}= 5

**Midpoints**

**Midpoints** are the points that lie directly in the middle of two other points. Let’s say that B is the midpoint of A and C. Then if there is a straight line between A and C then the point B **bisects** the line.

If A has the coordinates (x_1, y_1) and point C has coordinates (x_2, y_2) then the midpoint is

\text{Midpoint} = \left( \dfrac{x_1+x_2}{2}, \dfrac{y_1 + y_2}{2} \right) .

We can think of the midpoint as the average point between A and C

The** distance to the midpoint** is, intuitively

\text{Distance }AB=\dfrac{1}{2} \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} =\text{Distance }BC

or

\text{Distance }AB=\sqrt{ \left( x_1 - \dfrac{x_2+x_1}{2}\right)^2+\left( y_1- \dfrac{y_2+y_1}{2}\right)^2} =\text{Distance }BC

**Example 1: Finding a Midpoint**

Find the midpoint between the points (3,8) and (15, 16)

**[2 marks]**

The midpoint

\text{Midpoint}= \left( \dfrac{3+15}{2}, \dfrac{8+16}{2} \right) = (9, 12)

**Example 2: Finding the Distance Between Two Points**

Find the distance between the points (17,9) and (11,17)

**[2 marks]**

Using Pythagoras for Cartesian coordinates

\text{Distance}= \sqrt{(17-11)^2+(9-17)^2},

\text{Distance}= \sqrt{(6)^2+(-8)^2}= \sqrt{36+64},

\text{Distance}= \sqrt{100}=10.

The distance cannot be -10, as distance is a positive value.

**Example 3: Finding the Distance To a Midpoint**

Find the midpoint between the points (5,6) and (13,14), and the distance between (5,6) and the midpoint.

**[4 marks]**

The midpoint

\text{Midpoint}= \left( \dfrac{5+13}{2}, \dfrac{6+14}{2} \right) = (9, 10)

Using Pythagoras for Cartesian coordinates

\text{Distance}= \sqrt{(9-5)^2+(10-6)^2},

\text{Distance}= \sqrt{(4)^2+(4)^2}= 4\sqrt{2}.

The distance cannot be - 4\sqrt{2}, as distance is a positive value.

## Coordinates and Midpoints Example Questions

**Question 1: **Find the distance between the points (5, 12) and (60,60)

**[2 marks]**

We use Pythagoras’s theorem for Cartesian products.

\text{Distance}= \sqrt{(60-5)^2+(60-12)^2}

\text{Distance}= \sqrt{55^2+48^2}

\text{Distance}= \sqrt{3025+2304}= \sqrt{5329}=73

The distance between the two points is 73

**Question 2: **Find the midpoint between the points (4, 16) and (24,-16)

**[2 marks]**

The midpoint is

\text{Midpoint}= \left( \dfrac{4+24}{2}, \dfrac{16+ (-16)}{2} \right)

\text{Midpoint}= (14,0)**Question 3: **Points A and B have a midpoint at (11, 43). Given that point A has coordinates (2,3)

a) Find the coordinates of point B.

**[3 marks]**

b) Find the distance between point A and point B.

**[2 marks]**

a) Let B = (b_x, b_y). Then we have

(11,43) = \left( \dfrac{2+b_x}{2}, \dfrac{3+b_y}{2} \right)

So for b_x,

11 = \dfrac{2+b_x}{2}

22 = 2+b_x

b_x=20.

For b_y,

43 = \dfrac{3+b_y}{2}

86 = 3 + b_y

b_y = 83

So B= (20, 83)

b) Using Pythagoras’s theorem for Cartesian products.

\text{Distance}= \sqrt{(20-2)^2+(83-3)^2}

\text{Distance}= \sqrt{18^2+80^2}

\text{Distance}= \sqrt{324+6400}= \sqrt(6724)=82

The distance between the two points is 82

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