Coordinates and Midpoints

GCSELevel 4-5Level 6-7Cambridge iGCSE

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.

Level 4-5GCSECambridge iGCSE

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

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

Level 4-5GCSECambridge iGCSE

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

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

Level 6-7GCSECambridge iGCSE

Example 1: Finding a Midpoint

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

[2 marks]

The midpoint

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

Level 6-7GCSECambridge iGCSE

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.

Level 4-5GCSECambridge iGCSE

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.

Level 6-7GCSECambridge iGCSE

Coordinates and Midpoints Example Questions

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 $73$

The midpoint is

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

$\text{Midpoint}= (14,0)$

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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