Midpoint of a line segment

Let A(x_1,y_1) and B(x_2,y_2) be the coordinates of a line segment AB. We have:

    \[\boxed{\textrm{\textbf{MIDPOINT FORMULA}:=}\Big(\frac{x_1 +x_2}{2},\frac{y_1 + y_2}{2}\Big)}\]

Simple, right? Very simple. Straight to examples.

Example 1

Find the midpoint of the line segment joining A(-1,2) and B(3,-8).

    \[\boxed{\textrm{\textbf{midpoint=}}\Big(\frac{-1+3}{2},\frac{2+(-8)}{2}\Big)=\Big(\frac{2}{2},\frac{-6}{2}\Big)=(1,-3)}\]

Example 2

Find the midpoint of A(-2,-4) and B(-1,-5).

    \[\boxed{\textrm{\textbf{midpoint=}}\Big(\frac{-2+-1}{2},\frac{-4+-5}{2}\Big)=\Big(\frac{-3}{2},\frac{-9}{2}\Big)}\]

Example 3

Find the midpoint of the line segment joining A(\frac{1}{3},2) and B(-\frac{1}{5},\frac{3}{4})

    \[\boxed{\textrm{\textbf{midpoint=}}\Big(\frac{1/3+(-1/5)}{2},\frac{2+(3/4)}{2}\Big)=\Big(\frac{1}{15},\frac{11}{8}\Big)}\]

Example 4

Find the midpoint of the line segment joining A(-\frac{5}{2},\frac{3}{5}) and B(-\frac{7}{2},\frac{7}{5})

    \[\boxed{\textrm{\textbf{midpoint=}}\Big(\frac{-5/2+(-7/2)}{2},\frac{3/5+7/5}{2}\Big)=(-3,1)}\]

Example 5

Find the midpoint of the line segment joining A(\frac{11}{6},\frac{13}{3}) and B(\frac{5}{3},\frac{7}{6})

    \[\boxed{\textrm{\textbf{midpoint=}}\Big(\frac{11/6+5/3}{2},\frac{13/3+7/6}{2}\Big)=\Big(\frac{7}{4},\frac{11}{4}\Big)}\]

Example 6

A diameter of a circle has endpoints located at A(-3,6) and B(3,-6). Find the coordinates of the center of the circle.

The center of a circle is at the midpoint of any of its diameters, so we simply have to find the midpoint of A(-3,6) and B(3,-6) in this case; it is:

    \[\Big(\frac{-3+3}{2},\frac{6+-6}{2}\Big)=(0,0).\]

So this circle is centered at (0,0), the origin.

Example 7

How many line segments have midpoints at the origin (0,0)?

The answer is infinite. In fact, for any number a\neq 0, the points (a,a) and (-a,-a) all have midpoints at

    \[\Big(\frac{a+-a}{2},\frac{a+-a}{2}\Big)=(0,0),\]

which is the origin. In general, this also works for any point other than the origin: once the midpoint is specified, there are infinite line segments that share that midpoint. On the other hand, a given line segment will always have a unique midpoint.

In the terminology of functions (don’t worry about this yet if you’re still in grade 10), the assignment

    \[A,B\mapsto\textrm{midpoint of}~A,B\]

is a function, while its inverse is not.

Example 8

Line segment AB has midpoint at M(1,3). If A is the point (-2,7), find the coordinates of B.

Let B be the point with coordinates (x,y). Since the midpoint of A(-2,7) and B(x,y) is M(1,3), we have, by the midpoint formula, that

    \[(1,3)=\Big(\frac{-2+x}{2},\frac{7+y}{2}\Big).\]

The x-coordinates must be equal, so 1=\frac{-2+x}{2}; similarly, the y-coordinates must be equal, so 3=\frac{7+y}{2}.

    \begin{align*} \frac{-2+x}{2}&=1\\ -2+x&=1\times 2\quad\textrm{(clear fractions using the LCD)}\\ -2+x&=2\\ x&=4\\ \vdots&\\ \frac{7+y}{2}&=3\\ 7+y&=3\times 2\quad\textrm{(clear fractions using the LCD)}\\ 7+y&=6\\ y&=-1 \end{align*}

Therefore, B is the point (x,y)=(4,-1).

Example 9

Prove that the midpoint of (1,-1) and (3,-5) lies on the line y=\frac{1}{2}x-4.

We first calculate the midpoint of the given points (1,-1) and (3,-5); it is:

    \[\Big(\frac{1+3}{2},\frac{-1+-5}{2}\Big)=(2,-3).\]

Next, we check whether the midpoint we obtained, namely (2,-3), is on the given line y=\frac{1}{2}x-4. To do this, let’s substitute the value of x (that is, 2) into the given equation and see if we’ll get the value of y (that is, -3):

    \begin{align*} y&=\frac{1}{2}x-4\\ &=\frac{1}{2}(2)-4\quad\textrm{[we've replaced x with 2]}\\ &=1-4\\ &=-3 \end{align*}

Since we obtained the desired value of y, we can conclude that the midpoint lies on the line y=\frac{1}{2}x-4.

Example 10

If the midpoint of (x,y) and (y,-x) is (1,3), determine the values of x and y.

We have, by the midpoint formula, that:

    \[(1,3)=\Big(\frac{x+y}{2},\frac{y+-x}{2}\Big).\]

This in turn leads to two separate equations:

    \[1=\frac{x+y}{2},\quad\textrm{and}\quad 3=\frac{y+-x}{2}.\]

Clear fractions to obtain

    \[2=x+y\quad\textrm{and}\quad 6=y+-x.\]

If we re-arrange a bit, we see that this is a linear system:

    \begin{align*} x+y&=2\\ -x+y&=6 \end{align*}


Adding the two equations: 2y=8\implies y=4. Substituting y=4 in any of the two equations gives x=-2. So we conclude that x=-2 and y=4.

Exercises for the reader

  1. Derive the midpoint formula; that is, prove that the midpoint of A(x_1,y_1) and B(x_2,y_2) is \Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big).
  2. An idea from statistics: Let M\Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big) be the midpoint of A(x_1,y_1) and B(x_2,y_2). Define the lower midpoint L to be the midpoint of A and M, and the upper midpoint U to be the midpoint of M and B. Prove that:

        \[L=\Big(\frac{3x_1+x_2}{4},\frac{3y_1+y_2}{4}\Big);~U=\Big(\frac{x_1+3x_2}{4},\frac{y_1+3y_2}{4}\Big).\]

  3. In furtherance of exercise 2 above, prove that the midpoint of L and U is M, the midpoint of A and B. (This is expected.)

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