# Altitudes and right bisectors

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

Altitudes can also be drawn from vertex or vertex . 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 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 is the right bisector of the red line segment .

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 and are the slopes of two perpendicular lines, then:

Don’t forget this.

#### Example 1

has vertices located at . Find the equation of the altitude from .

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

The slope of is:

Since the altitude is perpendicular to , the slope of is then . So we can write the equation of in the form where :

So the equation of the altitude through is .

#### Example 2

has vertices located at . Find the equation of the altitude from .

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

Thus, the slope of the altitude from will be . We can then write its equation in the form where :

The equation of the altitude from is .

#### Example 3

has vertices located at . Find the equation of the altitude from .

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

Since is perpendicular to the -axis at the point , that’s its equation.

#### Example 4

Find the coordinates of the orthocenter of whose vertices are located at .

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:

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

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

Therefore, the orthocenter is located at .

#### Example 5

has vertices located at . Find the equation of the right bisector of side .

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

We first find the slope of ; it is:

Since the right bisector is perpendicular to , its slope will be the negative reciprocal of ; that is, . So, the equation of can be written in the form

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

Thus the point in the diagram has coordinates . Using this in the equation enables us to determine the value of :

And so is the equation of the right bisector of side .

#### Example 6

has vertices located at . Find the equation of the right bisector of side .

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

The slope of side is . The two things we need are:

The equation of is then . At point :

So the bisector has equation .

#### Example 7

has vertices located at . Find the equation of the right bisector of side .

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

The slope of is , so the slope of its right bisector is , and its equation is then

Since passes through the midpoint of (i.e. the point with coordinates ), we have that:

Therefore, the equation of the right bisector of side is .

#### Example 8

has vertices located at . 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:

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:

We conclude that the circumcenter is at . 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 and an external point , we can determine the perpendicular distance from 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 to the line .

Observe that the given point is not on the line (verify this by substituting the coordinates of into the left side of the equation). So we can draw a perpendicular line from to . Let the foot of this perpendicular be . Since lies on the line , its coordinates satisfy the equation, and so

(1)

Next, the slope of the line is . The slope of the given line is (re-arrange the equation in the form to see this). Since is perpendicular to this line, the product of their slopes is :

Re-arranging, we obtain:

(2)

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

is the point . Therefore, the distance from to is:

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

#### Example 10

Suppose that a line has equation and that a point is not on the line. PROVE that the perpendicular distance from to the line can be given by:

Following the procedure of our previous example, we let the foot of the perpendicular from to the line be . Since lies on the line , its coordinates satisfy the equation, and so

(3)

We need a second equation. To this end, note that the slope of is . Further, the slope of the line is . Since is perpendicular to the line, the product of their slopes is . Thus:

So the second equation is:

(4)

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

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

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

So is the point . Finally, we calculate the distance :

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 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 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.

1. has vertices at . Find the equations of all three altitudes. Hence determine its orthocenter.
2. has vertices located at . Find the equations of all three right bisectors of the sides of . Hence find the coordinates of its circumcenter.
3. PROVE that the equation of the right bisector joining the points and can be given by .
4. Let be the vertices of a triangle. PROVE that the equation of the altitude from can be given by . [Notice that the right bisector equation can be reobtained via the substitutions and . This says something about the relationship between an altitude and a right bisector.]
5. 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.]
6. Consider shown below (don’t assume that it’s isosceles; it’s been drawn that way for convenience):

If the altitudes and are equal in length, PROVE that and that .

7. Assume now that the diagram in the preceding question represents an isosceles triangle. If , PROVE that the triangle becomes equilateral.
8. are the vertices of . The “foot” of the altitude from to is at . PROVE that:
• the coordinates of are given by:

• the length of the altitude is:

• the area of is: