Altitudes and right bisectors – High School Geometry

In a triangle, an altitude is a perpendicular line drawn from one vertex to the opposite side. For example, if we consider $\triangle PQR$ shown below, the altitude from vertex $R$ is the red line segment $RN$ :

Altitudes can also be drawn from vertex $P$ or vertex $Q$ . The three altitudes meet at a point called the orthocenter.

Unlike a median, an altitude need not go through the midpoint of the opposite side. Moreover, it can happen that the altitude does not meet the opposite side “directly”, unless that side is “produced” or “extended”. Such an altitude is called an external altitude, while the regular ones are called internal.

A right bisector is a line that bisects (divides into two equal parts) a line segment and also makes an angle of $90^{\circ}$ with it. In particular, we can draw right bisectors for the three sides of a triangle, and they will meet at a point called the circumcenter.

In the above diagram, the blue line $RS$ is the right bisector of the red line segment $PQ$ .

It may be useful to visualize the relationship between a median (RM), an altitude (RN), and a right bisector (MU) in a single diagram:

Equations of altitudes and right bisectors

Since altitudes and right bisectors are perpendicular to the opposite side, finding their equations relies on an important property about slopes: if two lines are perpendicular, then the product of their slopes is NEGATIVE ONE. In other words, if $m_1$ and $m_2$ are the slopes of two perpendicular lines, then:

$m_1\times m_2=-1,\quad\textrm{or}~m_2=\frac{-1}{m_1}.$

Don’t forget this.

Example 1

$\triangle PQR$ has vertices located at $P(-4,0),~Q(4,0),~R(1,4)$ . Find the equation of the altitude from $P$ .

A diagram showing the altitude from $P$ (red line segment $PA$ ):

The slope of $QR$ is:

$\frac{0-4}{4-1}=-\frac{4}{3}.$

Since the altitude $PA$ is perpendicular to $QR$ , the slope of $PA$ is then $-1/(-4/3)=\frac{3}{4}$ . So we can write the equation of $PA$ in the form $y=mx+b$ where $m=\frac{3}{4}$ :

$\begin{equation*} \begin{split} y&=mx+b\\ y&=\frac{3}{4}x+b\quad\textrm{because}~m=\frac{3}{4}\\ 0&=\frac{3}{4}(-4)+b\quad\textrm{using point P(-4,0)}\\ 0&=-3+b\\ 3&=b\\ &\vdots\cdots\vdots\cdots\vdots\\ y&=\frac{3}{4}x+3 \end{split} \end{equation}$

So the equation of the altitude through $P$ is $y=\frac{3}{4}x+3$ .

Example 2

$\triangle PQR$ has vertices located at $P(-4,0),~Q(4,0),~R(1,4)$ . Find the equation of the altitude from $Q$ .

The altitude from $Q$ will be perpendicular to $PR$ . So we first find the slope of $PR$ ; it is:

$\frac{4-0}{1-(-4)}=\frac{4}{5}.$

Thus, the slope of the altitude from $Q$ will be $-\frac{5}{4}$ . We can then write its equation in the form $y=mx+b$ where $m=-\frac{5}{4}$ :

$\begin{equation*} \begin{split} y&=mx+b\\ y&=-\frac{5}{4}x+b\\ 0&=-\frac{5}{4}(4)+b\quad\textrm{using point}~Q(4,0)\\ 0&=-5+b\\ 5&=b\\ &\vdots\cdots\vdots\cdots\vdots\\ y&=-\frac{5}{4}x+5 \end{split} \end{equation}$

The equation of the altitude from $Q$ is $y=-\frac{5}{4}x+5$ .

Example 3

$\triangle PQR$ has vertices located at $P(-4,0),~Q(4,0),~R(1,4)$ . Find the equation of the altitude from $R$ .

The altitude from $R$ is the blue line segment $RN$ shown in the diagram below:

Since $RN$ is perpendicular to the $x$ -axis at the point $x=1$ , that’s its equation.

Example 4

Find the coordinates of the orthocenter of $\triangle PQR$ whose vertices are located at $P(-4,0),~Q(4,0),~R(1,4)$ .

As you can observe, this is the same triangle we’ve been dealing with since the first example. So we already have the equations of the three altitudes; they are:

$\begin{equation*} \begin{split} y&=\frac{3}{4}x+3\quad\textrm{altitude from P}\\ y&=-\frac{5}{4}x+5\quad\textrm{altitude from Q}\\ x&=1\quad\textrm{altitude from R} \end{split} \end{equation}$

All that remains is to solve the above system. Already we have $x=1$ , but let’s see if we can reobtain this value by solving the first two equations:

$\begin{equation*} \begin{split} y&=\frac{3}{4}x+3\\ y&=-\frac{5}{4}x+5\\ &\vdots\cdots\vdots\\ \frac{3}{4}x+3&=-\frac{5}{4}x+5\\ \frac{3}{4}x+\frac{5}{4}x&=5-3\\ \frac{8}{4}x&=2\\ 2x&=2\\ x&=1 \end{split} \end{equation}$

Yes, we got $x=1$ . It was probably unnecessary to do this, but just to be double sure that the system is consistent. With $x=1$ , the value of $y$ is:

$y=\frac{3}{4}x+3=\frac{3}{4}(1)+3=\frac{3}{4}+\frac{12}{4}=\frac{15}{4}$

Therefore, the orthocenter is located at $(1,\frac{15}{4})$ .

Example 5

$\triangle ABC$ has vertices located at $A(0,0),~B(2,3),~C(1,-2)$ . Find the equation of the right bisector of side $AB$ .

The situation is pictured below, where the right bisector of $AB$ is the dashed line $NS$ .

We first find the slope of $AB$ ; it is:

$\frac{3-0}{2-0}=\frac{3}{2}.$

Since the right bisector $NS$ is perpendicular to $AB$ , its slope will be the negative reciprocal of $\frac{3}{2}$ ; that is, $\frac{-2}{3}$ . So, the equation of $NS$ can be written in the form

$y=\frac{-2}{3}x+b.$

We now need to find $b$ . To do so, we use the fact that the right bisector passes through the midpoint of $AB$ :

$\boxed{\textrm{\textbf{midpoint of AB:=}}\Big(\frac{0+2}{2},\frac{0+3}{2}\Big)=\Big(1,\frac{3}{2}\Big)}$

Thus the point $N$ in the diagram has coordinates $(1,3/2)$ . Using this in the equation $y=-\frac{2}{3}x+b$ enables us to determine the value of $b$ :

$\begin{equation*} \begin{split} y&=-\frac{2}{3}x+b\\ \frac{3}{2}&=-\frac{2}{3}(1)+b\\ \frac{3}{2}+\frac{2}{3}&=b\\ \frac{3\times 3}{2\times 3}+\frac{2\times 2}{3\times 2}&=b\quad\textrm{common denominator}\\ \frac{9}{6}+\frac{4}{6}&=b\\ \frac{13}{6}&=b \end{split} \end{equation}$

And so $y=-\frac{2}{3}x+\frac{13}{6}$ is the equation of the right bisector of side $AB$ .

Example 6

$\triangle ABC$ has vertices located at $A(0,0),~B(2,3),~C(1,-2)$ . Find the equation of the right bisector of side $AC$ .

In the diagram below, the right bisector of side $AC$ is the blue line $MR$ .

The slope of side $AC$ is $\frac{-2-0}{1-0}=-2$ . The two things we need are:

$\boxed{\textrm{\textbf{slope of MR:=}}\frac{1}{2};\quad\textrm{ \textbf{midpoint of AC (point M):=}}\Big(\frac{1}{2},-1\Big)}$

The equation of $MR$ is then $y=\frac{1}{2}x+b$ . At point $M(\frac{1}{2},-1)$ :

$\begin{equation*} \begin{split} -1&=\frac{1}{2}\Big(\frac{1}{2}\Big)+b\\ -1&=\frac{1}{4}+b\\ -1-\frac{1}{4}&=b\\ \frac{-1\times 4}{1\times 4}-\frac{1}{4}&=b\quad\textrm{common denominator}\\ \frac{-4}{4}-\frac{1}{4}&=b\\ -\frac{5}{4}&=b \end{split} \end{equation}$

So the bisector $MR$ has equation $y=\frac{1}{2}x-\frac{5}{4}$ .

Example 7

$\triangle ABC$ has vertices located at $A(0,0),~B(2,3),~C(1,-2)$ . Find the equation of the right bisector of side $BC$ .

The right bisector of side $BC$ is the dashed line $UV$ shown below:

The slope of $BC$ is $5$ , so the slope of its right bisector $UV$ is $-\frac{1}{5}$ , and its equation is then

$y=-\frac{1}{5}x+b.$

Since $UV$ passes through the midpoint of $BC$ (i.e. the point $U$ with coordinates $(\frac{3}{2},\frac{1}{2})$ ), we have that:

$\begin{equation*} \begin{split} \frac{1}{2}&=-\frac{1}{5}\Big(\frac{3}{2}\Big)+b\\ \frac{1}{2}&=-\frac{3}{10}+b\\ \frac{1}{2}+\frac{3}{10}&=b\\ \frac{1\times 5}{2\times 5}+\frac{3}{10}&=b\quad\textrm{common denominator}\\ \frac{5}{10}+\frac{3}{10}&=b\\ \frac{8}{10}&=b\\ \frac{4}{5}&=b\quad\textrm{reduce to lowest terms} \end{split} \end{equation}$

Therefore, the equation of the right bisector of side $BC$ is $y=-\frac{1}{5}x+\frac{4}{5}$ .

Example 8

$\triangle ABC$ has vertices located at $A(0,0),~B(2,3),~C(1,-2)$ . Find its circumcenter.

The circumcenter, as you’ll recall, is the meeting point of the three right bisectors of a triangle. In the present case, we’ve already determined the equations of the three right bisectors:

$\begin{equation*} \begin{split} y&=-\frac{2}{3}x+\frac{13}{6}\quad\longrightarrow\textrm{bisector of AB}\\ y&=\frac{1}{2}x-\frac{5}{4}\quad\longrightarrow\textrm{bisector of AC}\\ y&=-\frac{1}{5}x+\frac{4}{5}\quad\longrightarrow\textrm{bisector BC} \end{split} \end{equation}$

It now remains to solve the three equations simultaneously. We take any two, and use the third one to test for consistency (you’ll not have to do this in practice; any two will do). Let’s solve the first two equations:

$\begin{equation*} \begin{split} y&=-\frac{2}{3}x+\frac{13}{6}\\ y&=\frac{1}{2}x-\frac{5}{4}\\ &\vdots\cdots\vdots\\ -\frac{2}{3}x+\frac{13}{6}&=\frac{1}{2}x-\frac{5}{4}\\ -\frac{2}{3}x-\frac{1}{2}x&=-\frac{5}{4}-\frac{13}{6}\\ -\frac{4}{6}x-\frac{3}{6}x&=-\frac{15}{12}-\frac{26}{12}\\ -\frac{7}{6}x&=-\frac{41}{12}\\ x&=\frac{\frac{41}{12}}{\frac{7}{6}}\\ x&=\frac{41}{12}\times\frac{6}{7}\\ x&=\frac{41}{14}\\ &\vdots\cdots\vdots\\ y&=-\frac{2}{3}x+\frac{13}{6}\\ y&=-\frac{2}{3}\Big(\frac{41}{14}\Big)+\frac{13}{6}\\ y&=-\frac{82}{42}+\frac{13}{6}\\ y&=-\frac{82}{42}+\frac{13\times 7}{6\times 7}\quad\textrm{common denominator}\\ y&=-\frac{82}{42}+\frac{91}{42}\\ y&=\frac{9}{42}\\ y&=\frac{3}{14}\\ &\vdots\cdots\vdots\\ &\vdots\cdots\vdots\\ &\vdots\cdots\vdots\\ &\textrm{use the third equation to check for consistency}\\ y&=\frac{1}{2}x-\frac{5}{4}\\ \frac{3}{14}&?=\frac{1}{2}\Big(\frac{41}{14}\Big)-\frac{5}{4}\\ \frac{3}{14}&?=\frac{41}{28}-\frac{5}{4}\\ \frac{3}{14}&?=\frac{41}{28}-\frac{5\times 7}{4\times 7}\quad\textrm{common denominator}\\ \frac{3}{14}&?=\frac{41}{28}-\frac{35}{28}\\ \frac{3}{14}&?=\frac{6}{28}\\ \frac{3}{14}&=\frac{3}{14}\quad\textrm{reduce to lowest terms}\\ \end{split} \end{equation}$

We conclude that the circumcenter is at $\Big(\frac{41}{14},\frac{3}{14}\Big)$ . See the diagram below (it shows that the circumcenter need not be inside the triangle).

Perpendicular distance from a point to a line

Given a line $Ax+By+C=0$ and an external point $P(x_1,y_1)$ , we can determine the perpendicular distance from $P$ to the line. This entails finding the “foot” of the perpendicular, and then using the distance formula.

Example 9

Find the perpendicular distance from the point $P(-1,2)$ to the line $x-y+2=0$ .

Observe that the given point $P(-1,2)$ is not on the line $x-y+2=0$ (verify this by substituting the coordinates of $P$ into the left side of the equation). So we can draw a perpendicular line from $P(-1,2)$ to $x-y+2=0$ . Let the foot of this perpendicular be $N(a,b)$ . Since $N(a,b)$ lies on the line $x-y+2=0$ , its coordinates satisfy the equation, and so

(1) $\begin{equation*} a-b+2=0 \end{equation*}$

Next, the slope of the line $PN$ is $\frac{b-2}{a--1}=\frac{b-2}{a+1}$ . The slope of the given line $x-y+2=0$ is $1$ (re-arrange the equation in the form $y=x+2$ to see this). Since $PN$ is perpendicular to this line, the product of their slopes is $-1$ :

$\frac{b-2}{a+1}\times 1=-1\implies b-2=-(a+1).$

Re-arranging, we obtain:

(2) $\begin{equation*} b=-a+1 \end{equation*}$

We now solve equations (1) and (2) simultaneously, using substitution method:

$\begin{equation*} \begin{split} a-b+2&=0\\ b&=-a+1\\ \vdots\cdots\vdots&=\vdots\cdots\vdots\\ a-(-a+1)+2&=0\\ a+a-1+2&=0\\ 2a+1&=0\\ 2a&=-1\\ a&=-\frac{1}{2}\\ \vdots\cdots\vdots&=\vdots\cdots\vdots\\ b&=-a+1\\ b&=-\Big(-\frac{1}{2}\Big)+1\\ b&=\frac{1}{2}+1\\ b&=\frac{3}{2} \end{split} \end{equation}$

$N$ is the point $(\frac{1}{2},\frac{3}{2})$ . Therefore, the distance from $P(-1,2)$ to $N(\frac{1}{2},\frac{3}{2})$ is:

$\sqrt{\Big(\frac{1}{2}--1\Big)^2+\Big(\frac{3}{2}-2\Big)^2}=\sqrt{\Big(\frac{3}{2}\Big)^2+\Big(\frac{-1}{2}\Big)^2}=\frac{\sqrt{10}}{2}.$

Let’s generalize the above procedure in our next example.

Example 10

Suppose that a line has equation $Ax+By+C=0$ and that a point $P(x_1,y_1)$ is not on the line. PROVE that the perpendicular distance from $P(x_1,y_1)$ to the line can be given by:

$\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}},$

where a negative sign in the final result can be ignored.

Following the procedure of our previous example, we let the foot of the perpendicular from $P(x_1,y_1)$ to the line be $N(a,b)$ . Since $N(a,b)$ lies on the line $Ax+By+C=0$ , its coordinates satisfy the equation, and so

(3) $\begin{equation*} Aa+Bb+C=0 \end{equation*}$

We need a second equation. To this end, note that the slope of $PN$ is $\frac{b-y_1}{a-x_1}$ . Further, the slope of the line $Ax+By+C=0$ is $-\frac{A}{B}$ . Since $PN$ is perpendicular to the line, the product of their slopes is $-1$ . Thus:

$\frac{b-y_1}{a-x_1}\times \Big(\frac{-A}{B}\Big)=-1\implies b=y_1+\frac{B}{A}\Big(a-x_1\Big)$

So the second equation is:

(4) $\begin{equation*} b=y_1+\frac{B}{A}\Big(a-x_1\Big) \end{equation*}$

Substitute the right side of equation (4) for $b$ in equation (3):

$Aa+B\Big(y_1+\frac{B}{A}\Big(a-x_1\Big)\Big)+C=0$

Multiply each term by $A$ to clear fractions and continue the simplification:

$\begin{equation*} \begin{split} A^2a+AB\Big(y_1+\frac{B}{A}\Big(a-x_1\Big)\Big)+AC&=0\\ A^2a+B\Big(Ay_1+B\Big(a-x_1\Big)\Big)+AC&=0\\ A^2a+B\Big(Ay_1+Ba-Bx_1\Big)+AC&=0\\ A^2a+BAy_1+B^2a-B^2x_1+AC&=0\\ (A^2+B^2)a+ABy_1-B^2x_1+AC&=0\\ (A^2+B^2)a&=B^2x_1-ABy_1-AC\\ a&=\frac{B^2x_1-ABy_1-AC}{A^2+B^2} \end{split} \end{equation}$

Using this expression for $a$ in equation (4) and simplifying, we obtain:

$b=\frac{A^2y_1-ABx_1-BC}{A^2+B^2}.$

So $N$ is the point $\Big(\frac{B^2x_1-ABy_1-AC}{A^2+B^2},\frac{A^2y_1-ABx_1-BC}{A^2+B^2}\Big)$ . Finally, we calculate the distance $PN$ :

$\begin{equation*} \begin{split} PN^2&=\Big(x_1-\Big(\frac{B^2x_1-ABy_1-AC}{A^2+B^2}\Big)\Big)^2+\Big(y_1-\Big(\frac{A^2y_1-ABx_1-BC}{A^2+B^2}\Big)\Big)^2\\ &=\Big(\frac{A^2x_1+ABy_1+AC}{A^2+B^2}\Big)^2+\Big(\frac{ABx_1+B^2y_1+BC}{A^2+B^2}\Big)^2\\ &=\frac{A^2}{(A^2+B^2)^2}\Big(Ax_1+By_1+C\Big)^2+\frac{B^2}{(A^2+B^2)^2}\Big(Ax_1+By_1+C\Big)^2\\ &=\Big(Ax_1+By_1+C\Big)^2\Big(\frac{A^2+B^2}{(A^2+B^2)^2}\Big)\\ &=\frac{(Ax_1+By_1+C)^2}{A^2+B^2}\\ &\vdots\cdots\vdots\\ PN&=\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}} \end{split} \end{equation}$

This proves the desired result.

Takeaway

In a triangle, an altitude originates from a vertex and is perpendicular to the opposite side. A right bisector divides a side equally and makes an angle of $90^{\circ}$ with it, but need not go through a vertex. A median originates from a vertex and bisects the opposite side, but doesn’t necessarily make a $90^{\circ}$ angle with it. If the triangle is equilateral, then all three (medians, altitudes, right bisectors) coincide.
Apart from medians, altitudes, and right bisectors, there are other important line segments connected to a triangle.

Tasks

$\triangle ABC$ has vertices at $A(-2,0),~B(4,0),~C(2,6)$ . Find the equations of all three altitudes. Hence determine its orthocenter.
$\triangle PQR$ has vertices located at $P(-1,-2),~Q(3,0),~R(2,2)$ . Find the equations of all three right bisectors of the sides of $\triangle PQR$ . Hence find the coordinates of its circumcenter.
PROVE that the equation of the right bisector joining the points $(x_1,y_1)$ and $(x_2,y_2)$ can be given by $2(y_2-y_1)y+2(x_2-x_1)x=(x_2^2-x_1^2)+(y_2^2-y_1^2)$ .
Let $(x_1,y_1),~(x_2,y_2),~(x_3,y_3)$ be the vertices of a triangle. PROVE that the equation of the altitude from $(x_1,y_1)$ can be given by $(y_3-y_2)y+(x_3-x_2)x=(y_3-y_2)y_1+(x_3-x_2)x_1$ . [Notice that the right bisector equation can be reobtained via the substitutions $x_1=\frac{x_2+x_3}{2}$ and $y_1=\frac{y_2+y_3}{2}$ . This says something about the relationship between an altitude and a right bisector.]
PROVE that if two altitudes of a triangle are equal in length, then the triangle must be isosceles. [The equal sides will be the two sides that contain the “feet” of the two equal altitudes.]
Consider $\triangle PQR$ shown below (don’t assume that it’s isosceles; it’s been drawn that way for convenience):

If the altitudes $PM$ and $RN$ are equal in length, PROVE that $\sin(\alpha_1)+\sin(\alpha_2)=1$ and that $\sin(\beta_1)+\sin(\beta_2)=1$ .
Assume now that the diagram in the preceding question represents an isosceles triangle. If $PM=RN$ , PROVE that the triangle becomes equilateral.
are the vertices of . The “foot” of the altitude from to is at . PROVE that:
- the coordinates of $N(a,b)$ are given by:
  $\begin{equation*} \begin{split} a&=\frac{x_1(x_3-x_2)^2+x_3(y_3-y_2)^2+(y_1-y_3)(y_3-y_2)(x_3-x_2)}{(x_3-x_2)^2+(y_3-y_2)^2}\\ b&=\frac{y_1(y_3-y_2)^2+y_3(x_3-x_2)^2+(x_1-x_3)(x_3-x_2)(y_3-y_2)}{(x_3-x_2)^2+(y_3-y_2)^2} \end{split} \end{equation}$
- the length of the altitude $AN$ is:
  $\begin{equation*} AN=\frac{y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)}{\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}} \end{equation}$
- the area of $\triangle ABC$ is:
  $\begin{equation*} \Delta=\frac{y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)}{2}. \end{equation}$