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Stage 5.1-2

# 3.01 The midpoint of a segment

Lesson

#### Exploration

Explore this applet demonstrating the midpoint between two points.

What connections exist between the endpoints of a line segment and the midpoint?

The midpoint of a line segment is a point exactly halfway along the segment. That is, the distance from the midpoint to both of the endpoints is the same.

The midpoint of any two points has coordinates that are exactly halfway between the $x$x-values and halfway between the $y$y-values. This means we can find the average of the two given $x$x-coordinates to find the $x$x-coordinate of the midpoint, and likewise the average of the two $y$y-coordinates will give us the $y$y-coordinate of the midpoint.

Finding the midpoint

So for points $A$A$\left(x_1,y_1\right)$(x1,y1) and $B$B$\left(x_2,y_2\right)$(x2,y2) the midpoint will be:

$M$M$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$(x1+x22,y1+y22)

Think of it as averaging the $x$x and $y$y-values of the endpoints.

### Visual of the midpoint

Find the midpoint $M$M$\left(x_m,y_m\right)$(xm,ym) between $A$A$\left(-5,12\right)$(5,12) and B$\left(3,4\right)$(3,4)

Think: The midpoint will be the halfway point of the $x$x and $y$y-values.  We can find it by first finding the average of the $x$x and $y$y-values of the endpoints.

Do: The $x$x-value of the midpoint will be the average of $x_1=-5$x1=5 and $x_2=3$x2=3

 $x_m$xm​ $=$= $\frac{x_1+x_2}{2}$x1​+x2​2​ $=$= $\frac{-5+3}{2}$−5+32​ Substitute in the given values $=$= $\frac{-2}{2}$−22​ Evaluate the sum in the numerator $=$= $-1$−1 Evaluate the quotient

So the $x$x-value of the midpoint is $x_m=-1$xm=1.

The $y$y-value of the midpoint will be the average of $y_1=12$y1=12 and $y_2=4$y2=4

 $y_m$ym​ $=$= $\frac{y_1+y_2}{2}$y1​+y2​2​ $=$= $\frac{12+4}{2}$12+42​ Substitute in the given values $=$= $\frac{16}{2}$162​ Evaluate the sum in the numerator $=$= $8$8 Evaluate the quotient

The $y$y-value of the midpoint is $y_m=8$ym=8.

This means the midpoint has coordinates $\left(-1,8\right)$(1,8).

Reflect: Does it matter which point we choose to be $\left(x_1,y_1\right)$(x1,y1) and which is $\left(x_2,y_2\right)$(x2,y2)?

By convention, we work from left to right so the point with the lower $x$x-value is generally considered to be the point corresponding to $\left(x_1,y_1\right)$(x1,y1), but obviously if we went from right to left we should still end up with the same midpoint. Let's check!

We could have calculated the midpoints as follows:

$x_m=\frac{3+\left(-5\right)}{2}$xm=3+(5)2 this simplifies to $\frac{-2}{2}=-1$22=1. Which is the value we found in the first place.

Similarly for the $y$y-value:

$y_m=\frac{4+12}{2}$ym=4+122 which simplifies to $\frac{16}{2}=8$162=8, again this is the value we found previously.

So it doesn't matter which point we choose to be $\left(x_1,y_1\right)$(x1,y1).

When we are dealing with negative values you may find it easier choosing one over the other.

### Working backwards to find an endpoint

What if we are given the midpoint of a segment, and one endpoints points of the segment? How can we reverse our steps above to find the other endpoint?

Suppose that the point $M$M$\left(2,-5\right)$(2,5) is the midpoint of $A$A$\left(-4,3\right)$(4,3) and $B$B. We want to find the coordinates of $B$B$\left(x,y\right)$(x,y).

Think: It would be helpful to sketch what this information looks like on the number plane, so we can anticipate where $B$B should lie.

Now to find $B$B, consider what we would usually do to find the coordinates of the midpoint.

We would find the average of the $x$x-values of $A$A and $B$B:

$\frac{-4+x}{2}$4+x2

and the average of the $y$y-values of $A$A and $B$B:

$\frac{3+y}{2}$3+y2

In this scenario, we are given the values of the midpoint. That is:

$\left(2,-5\right)=\left(\frac{-4+x}{2},\frac{3+y}{2}\right)$(2,5)=(4+x2,3+y2)

which gives us

$\frac{-4+x}{2}=2$4+x2=2, and $\frac{3+y}{2}=-5$3+y2=5

Do: Solve each equation for its $x$x and $y$y-value:

 $\frac{-4+x}{2}$−4+x2​ $=$= $2$2 $-4+x$−4+x $=$= $4$4 $-4+4+x$−4+4+x $=$= $4+4$4+4 $x$x $=$= $8$8
 $\frac{3+y}{2}$3+y2​ $=$= $-5$−5 $3+y$3+y $=$= $-10$−10 $3-3+y$3−3+y $=$= $-10-3$−10−3 $y$y $=$= $-13$−13

This means that $B$B has coordinates $\left(8,-13\right)$(8,13).

Reflect: Thinking logically, we can see that the midpoint's $x$x-value is $6$6 more than the $x$x-value of the known point (Compare $2$2 to $-4$4), so we would expect the $x$x-value of the endpoint to be $6$6 more the midpoint, and $2+6=8$2+6=8 as expected.

Similarly the $y$y-value of the midpoint is $8$8 less than the midpoint (Compare $-5$5 to $3$3), so we would expect the y-value to be $8$8 less than the the midpoint, and as expected $-5-8=13$58=13

#### Practice questions

##### Question 1

$M$M is the midpoint of $A$A $\left(5,-6\right)$(5,6) and $B$B $\left(5,2\right)$(5,2).

1. Find the coordinates of $M$M.

$M$M is the point $\left(\editable{},\editable{}\right)$(,)

##### Question 2

$M$M is the midpoint of Point $A$A $\left(-10,-6\right)$(10,6) and Point $B$B $\left(4,10\right)$(4,10).

1. What is the $x$x-coordinate of $M$M?

2. What is the $y$y-coordinate of $M$M?

3. Hence, plot the $M$M on the axes below:

##### Question 3

If the midpoint of $A$A$\left(x,y\right)$(x,y) and $B$B$\left(10,3\right)$(10,3) is $M$M$\left(8,-1\right)$(8,1):

1. Find the value of $x$x.

2. Find the value of $y$y.

3. Hence, what are the coordinates of $A$A?

$A$A $\left(\editable{},\editable{}\right)$(,)

### Outcomes

#### MA5.1-6NA

determines the midpoint, gradient and length of an interval, and graphs linear relationships