Median of a triangle – High School Geometry

In a triangle $ABC$ , the median from vertex $A$ is a line segment from $A$ to the midpoint of side $BC$ (red line $AM$ in the diagram below).

Similarly, the median from vertex $B$ is the line segment from $B$ to the midpoint of $AC$ , and the median from vertex $C$ is a line segment from $C$ to the midpoint of $AB$ . The three medians of a triangle always intersect at a point called the centroid.

Since a median is a line segment, we can find its equation, its length, and its midpoint. We can also find the angle the median makes with the opposite side.

Equation of a median

Recall that the equation of a straight line can be given in the form $y=mx+b$ , where $m$ is the slope and $b$ is the $y$ -intercept. So, supposing we want to find the equation of the median $AM$ in the diagram above, then:

we first determine the coordinates of point $M$ using the midpoint formula (since $M$ is the midpoint of the opposite side $BC$ by definition);
$\boxed{\textrm{\textbf{MIDPOINT FORMULA:=}}\Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big)}$
we find the slope of $AM$ using the slope formula (call this slope $m$ );
$\boxed{\textrm{\textbf{SLOPE FORMULA:=}}\frac{y_2-y_1}{x_2-x_1}}$
we substitute the slope $m$ and the coordinates of $M$ (or $A$ ) into the straight line equation $y=mx+b$ to find $b$ ;
we substitute $b$ back into $y=mx+b$ , which will then be the desired equation.

So long as you’re comfortable with finding equations of lines, you won’t have any issues with equations of medians.

Example 1

$\triangle PQR$ has vertices at $P(-4,0),~Q(6,0),~R(0,4)$ . Find the equation of the median from $R$ .

The median from vertex $R$ is the red line segment $RN$ shown above. Following the steps we outlined previously, we first find the midpoint of $PQ$ ; it is:

$\Big(\frac{-4+6}{2},\frac{0+0}{2}\Big)=(1,0).$

Let’s label this midpoint as $N(1,0)$ . Next, we find the slope of $RN$ ; it is:

$\frac{0-4}{1-0}=\frac{-4}{1}=-4=m,~\textrm{say}.$

We can now write $y=-4x+b$ , and then determine the value of $b$ using the coordinates of $N$ or $R$ . Let’s use $R(0,4)$ :

$\begin{equation*} \begin{split} y&=-4x+b\\ 4&=-4(0)+b,\quad\textrm{since}~x=0,y=4~\textrm{at}~R(0,4)\\ 4&=0+b\\ 4&=b. \end{split} \end{equation}$

Finally, in $y=-4x+b$ , we replace the $b$ by $4$ . Therefore, the desired equation is $y=-4x+4$ . (Notice that we could have saved a little time by observing that the point $R(0,4)$ is on the $y$ -axis, and so $b=4$ follows immediately.)

Example 2

$\triangle PQR$ has vertices at $P(-4,0),~Q(6,0),~R(0,4)$ . Find the equation of the median from $Q$ .

We first find the midpoint of $PR$ ; it is:

$\Big(\frac{0+-4}{2},\frac{4+0}{2}\Big)=(-2,2).$

Let’s label it as $M(-2,2)$ . Next, we find the slope of $QM$ ; it is:

$\frac{2-0}{-2-6}=\frac{2}{-8}=-\frac{1}{4}=m,~\textrm{say}.$

We can then write $y=-\frac{1}{4}x+b$ , and use the coordinates of $Q$ (or $M$ ) to determine the value of $b$ . Let’s use $M(-2,2)$ :

$\begin{equation*} \begin{split} y&=mx+b\\ 2&=-\frac{1}{4}(-2)+b\\ 2&=\frac{2}{4}+b\\ 2&=\frac{1}{2}+b\\ b&=2-\frac{1}{2}\\ b&=\frac{4}{2}-\frac{1}{2}\\ b&=\frac{3}{2} \end{split} \end{equation}$

So, the desired equation is $y=-\frac{1}{4}x+\frac{3}{2}$ .

Example 3

$\triangle PQR$ has vertices at $P(-4,0),~Q(6,0),~R(0,4)$ . Find the equation of the median from $P$ .

We first find the midpoint of $RQ$ ; it is:

$\Big(\frac{0+6}{2},\frac{4+0}{2}\Big)=(3,2).$

Let’s label it as $L(3,2)$ . Next, we find the slope of $PL$ ; it is:

$\frac{2-0}{3-(-4)}=\frac{2}{7}=m,~\textrm{say}.$

We can now write $y=\frac{2}{7}x+b$ , and use the coordinates of $P$ (or $L$ ) to determine $b$ . Let’s use $L(3,2)$ :

$\begin{equation*} \begin{split} y&=mx+b\\ 2&=\frac{2}{7}(3)+b\\ 2&=\frac{6}{7}+b\\ b&=2-\frac{6}{7}\\ b&=\frac{14}{7}-\frac{6}{7}\\ b&=\frac{8}{7} \end{split} \end{equation}$

So, the desired equation is $y=\frac{2}{7}x+\frac{8}{7}$ . (This can be written in other forms if one wishes.)

Example 4

$\triangle ABC$ has vertices located at points $A(-2,4),~B(4,2),~C(r,s)$ . The equation of the median through $B$ is $y=-\frac{1}{7}x+\frac{18}{7}$ , while the median through $C$ has equation $y=-x+4$ . Determine the values of $r$ and $s$ .

Since we have equations for two medians, it is possible to determine the coordinates of $C$ uniquely. Let’s draw a diagram first (we do not know precisely where the point $C$ will be, so we just insert it somewhere).

Since $C(r,s)$ lies on the median $y=-x+4$ , its coordinates satisfy this equation, so

(1) $\begin{equation*} s=-r+4 \end{equation*}$

Also, the midpoint of $AC$ , namely

$\Big(\frac{r+-2}{2},\frac{s+4}{2}\Big)=\Big(\frac{r-2}{2},\frac{s+4}{2}\Big)$

lies on the median through $B$ (equation: $y=-\frac{1}{7}x+\frac{18}{7}$ ), so we obtain our second equation:

(2) $\begin{equation*} \frac{s+4}{2}=-\frac{1}{7}\Big(\frac{r-2}{2}\Big)+\frac{18}{7} \end{equation*}$

The lowest common denominator in equation (2) is $14$ . We clear the fractions by multiplying both sides of the equation by $14$ :

$14\times\Big(\frac{s+4}{2}\Big)=14\times\Big(-\frac{1}{7}\Big(\frac{r-2}{2}\Big)+\frac{18}{7}\Big)$

That is,

$7s+28=-r+2+36,\quad\implies r=10-7s.$

Substitute $10-7s$ for $r$ in equation (1):

$\begin{equation*} \begin{split} s&=-r+4\\ s&=-(10-7s)+4\\ s&=-10+7s+4\\ s-7s&=-10+4\\ -6s&=-6\\ s&=1 \end{split} \end{equation}$

Since $r=10-7s$ , we obtain $r=10-7\times 1=10-7=3$ . So $r=3$ and $s=1$ .

Example 5

$\triangle RST$ has vertices located at $R(-2,2),~S(4,6),~T(a,b)$ . If the equation of the median through $T$ is $y=4x$ , determine possible values for $a$ and $b$ . How many possibilities are there?

Imagine that we’ve represented the points in the diagram below:

Since the median $TM$ lies on the line $y=4x$ , the coordinates of point $T(a,b)$ must satisfy this equation, and so

$b=4a.$

That’s all we’ve got from the given information.

It follows that the solution is not unique. Any ordered pair $(a,b)$ that satisfies $b=4a$ — except $(1,4)$ — will do. For example, $(0,0),(2,8),(-1,-4)$ are acceptable ordered pairs for $(a,b)$ . Thus, for the second part of the question, there are infinite possibilities.
It is essential to exclude the point $M(1,4)$ from our list of possible solutions. Since $M(1,4)$ is collinear with $R(-2,2)$ and $S(4,6)$ , we cannot have a triangle $RST$ if $T$ is chosen as the point $T(1,4)$ .

Length of a median

We can make use of the distance formula to calculate the length of a median.

$\boxed{\textrm{\textbf{DISTANCE FORMULA:=}}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

Example 6

$\triangle ABC$ has vertices located at $A(0,0),~B(10,0),~C(0,4)$ . Find the length of the median through $A$ .

We first find the midpoint of $BC$ ; it is:

$\Big(\frac{10+0}{2},\frac{0+4}{2}\Big)=(5,2).$

Let’s label this midpoint as $M(5,2)$ . The required median is line segment $AM$ . Using the distance formula, the length of $AM$ is:

$\sqrt{(5-0)^2+(2-0)^2}=\sqrt{29}.$

Example 7

$\triangle ABC$ has vertices located at $A(0,0),~B(10,0),~C(0,4)$ . Find the length of the median through $B$ .

Since we need the median through $B$ , we’ll first find the midpoint of $AC$ ; it is:

$\Big(\frac{0+0}{2},\frac{0+4}{2}\Big)=(0,2).$

Let’s label it as $N$ . Then, by the distance formula, the length of $BN$ is:

$\sqrt{(0-10)^2+(2-0)^2}=\sqrt{104}=\sqrt{4\times 26}=2\sqrt{26}.$

The length of the median through $B$ is $2\sqrt{26}\approx 10.20$ units.

Example 8

The vertices of a triangle $TUV$ are located at $T(-1,2),~U(5,4),~V(a,b)$ . If the length of the median through $V$ is $5$ units, determine possible values for $a$ and $b$ .

Since we’re dealing with the median through $V$ , we’ll first find the midpoint of $T$ and $U$ ; it is:

$\Big(\frac{-1+5}{2},\frac{2+4}{2}\Big)=(2,3).$

Let’s label this midpoint as $L(2,3)$ . The median is then $VL$ . Since its length is $5$ units, we have, by the distance formula, that:

$\sqrt{(a-2)^2+(b-3)^2}=5.$

Square both sides: $(a-2)^2+(b-3)^2=25$ . We’ve solved this type of question before during our post on the length of a line segment. In the present case, we can take $a=-2$ and $b=0$ . There are several other possibilities.

Centroid of a triangle

We’ll apply our knowledge of linear systems to find the centroid of a triangle — the point of intersection of the three medians.

Example 9

Find the coordinates of the centroid of $\triangle ABC$ , where the vertices are located at $A(-2,0),~B(4,-2),~C(2,2)$ .

The easiest approach to this question is to use the centroid formula (see the exercises at the end), but we don’t recommend this.

Another approach is to get a grid paper and a well-sharpened pencil. Then plot the points, draw the medians, and determine where they intersect. But there’s a limitation with this approach: if the point of intersection of the medians is not comprised of integer coordinates, then reading them off from the grid may be difficult.

What we’ll do here is to go through the derivation of the equations of the medians $AM$ and $CL$ as shown in the diagram above, then solve the resulting system of equations to find where they intersect.

Median $AM$ (orange-coloured line above) is (a section of) the $x$ -axis, so its equation is simple, short, straightforward:

(3) $\begin{equation*} y=0 \end{equation*}$

Next, we find the equation of $CL$ (green line above). Notice that $L$ is the point $L(1,-1)$ . Let’s use another form of the equation of a straight line:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

and then substitute the coordinates of $C(2,2)$ as $(x_1,y_1)$ and $L(1,-1)$ as $(x_2,y_2)$ . This way we obtain:

$\begin{equation*} \begin{split} \frac{y-y_1}{x-x_1}&=\frac{y_2-y_1}{x_2-x_1}\\ \frac{y-2}{x-2}&=\frac{-1-2}{1-2}\\ \frac{y-2}{x-2}&=\frac{-3}{-1}\\ \frac{y-2}{x-2}&=3\\ y-2&=3(x-2)\\ y&=3x-4 \end{split} \end{equation}$

So, the equation of median $CL$ is:

(4) $\begin{equation*} y=3x-4 \end{equation*}$

We now solve the linear system given by equations (3) and (4). From equation (3), $y=0$ . Substitute $y=0$ in equation (4):

$\begin{equation*} \begin{split} y&=3x-4\\ 0&=3x-4\\ 4&=3x\\ \frac{4}{3}&=x \end{split} \end{equation}$

Thus, the point of intersection of medians $AM$ and $CL$ is $(\frac{4}{3},0)$ . Put differently, the centroid is $(\frac{4}{3},0)$ . (It is also essential to ensure that median $BN$ also passes through this point, but we’ll leave that part.)

Midpoint of midpoint of midpoint

That English doesn’t sound right. Right? Well, you might be right. But let’s see what that might mean.

Example 10

Prove that the midpoint of a median always coincides with the midpoint of the midpoints of the other two sides.

Let’s use a diagram to clarify things. Consider $\triangle ABC$ pictured below:

Here, $P$ is the midpoint of $AB$ ; $Q$ is the midpoint of $AC$ ; and $R$ is the midpoint of $BC$ . We want to show that the midpoint of $PQ$ and the midpoint of $AR$ coincide. This is very simple.

Let’s find the coordinates of $P,Q,R$ :

$\begin{equation*} \begin{split} P&:=\Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big);\\ Q&:=\Big(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2}\Big);\\ R&:=\Big(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\Big). \end{split} \end{equation}$

The midpoint of $PQ$ is:

$\Big(\frac{\frac{x_1+x_2}{2}+\frac{x_1+x_3}{2}}{2},\frac{\frac{y_1+y_2}{2}+\frac{y_1+y_3}{2}}{2}\Big)=\Big(\frac{2x_1+x_2+x_3}{4},\frac{2y_1+y_2+y_3}{4}\Big).$

The midpoint of $AR$ is:

$\Big(\frac{x_1+\frac{x_2+x_3}{2}}{2},\frac{y_1+\frac{y_2+y_3}{2}}{2}\Big)=\Big(\frac{2x_1+x_2+x_3}{4},\frac{2y_1+y_2+y_3}{4}\Big).$

So we see that these two midpoints coincide. Congrats! You’ve proved a geometric property.

Takeaway

In a triangle, a median is a line segment connecting a vertex to the midpoint of the opposite side. A triangle has three medians, and the medians intersect at a point called the centroid. Finding the equation of a median is as simple as finding the equation of a regular straight line.

Tasks

$\triangle ABC$ has vertices located at $A(0,0),~B(1,4),~C(6,0)$ . Through which of the vertices should a median be drawn, so that the median is parallel to the line $2x+y+3=0$ ?
Find the centroid of the triangle whose vertices are $(-2,0),~(2,0),(0,2)$ .
The vertices of $\triangle ABC$ are $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ . Prove that the centroid of $\triangle ABC$ is located at $\Big(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\Big)$ .
[A simple way to obtain this result is to use the fact that the centroid divides the median in the ratio $2:1$ . But what if one is unaware of this fact? Then one has to resort to finding the equations of two medians, and solving the resulting linear system. We highly encourage anyone who has the time to spare, tenacity of spirit, and the tools and stationery to try the latter approach. It is an abundantly rewarding exercise. You won’t get many opportunities to solve a math problem where each line occupies the entire length of a $1$ m chalkboard. Do the drill and feel the thrill. Then thank us when you’re free.]
Determine an appropriate choice of coordinates for the vertices $A,B,C$ of $\triangle ABC$ for which its centroid will be the origin $(0,0)$ . Is the answer unique?
Prove that the three medians in an equilateral triangle are equal in length. (It happens that medians and altitudes coincide for equilateral triangles, and so they have lengths of the form $a\sqrt{3}$ , where $a$ is any non-zero real number.)
Prove that if the three medians in a triangle are equal in length, then the triangle is equilateral. (In view of the preceding exercise, it means that we can characterize equilateral triangles using medians.)
Prove that two of the three medians in an isosceles triangle are equal in length.
Prove that if two medians in a triangle are equal in length, then the triangle is isosceles. (In view of the preceding exercise, it follows that we can characterize isosceles triangles using medians.)
Find the angle the median from $A$ makes with side $BC$ , if the triangle $ABC$ has vertices at $A(-1,-1),~B(1,1),~C(3,-3)$ .
In $\triangle ABC$ , it is desired that the median from vertex $A$ and the median from vertex $B$ meet at $90^{\circ}$ . Determine an appropriate choice of coordinates for the vertices $A,B,C$ that will make this happen.