The auto center

Got any triangle you need to “fix”? Bring it to the orthocenter where we meet. So the three altitudes seem to speak.

Or so we speak.

The orthocenter — which was noted as auto center in this author’s jotter (to mimic the familiar term \cdots) — is the meeting point of the three altitudes of a triangle. In coordinate geometry, finding the orthocenter usually involves two main steps:

  1. find the equations of any two altitudes;
  2. solve the two equations simultaneously to determine their intersection point, which is then the orthocenter.

That’s exactly our aim in this post, just that we do it for a triangle with arbitrary vertices, and thereby obtain a general orthocenter formula. We’ll then solve examples using the formula as well as the standard procedure.

Formula

Let A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3) be the vertices of a triangle ABC. Then the coordintes of its orthocenter can be given by:
x=\frac{x_1x_2(y_2-y_1)+x_2x_3(y_3-y_2)+x_1x_3(y_1-y_3)-(y_2-y_1)(y_3-y_2)(y_1-y_3)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}

y=\frac{y_1y_2(x_1-x_2)+y_2y_3(x_2-x_3)+y_3y_1(x_3-x_1)-(x_1-x_2)(x_2-x_3)(x_3-x_1)}{y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)}

Note that the denominators in both expressions for x and y are the same (see the exercises at the end), so it is enough to calculate one of them.

Example 1

\triangle ABC has vertices at A(0,5),~B(-1,2),~C(3,0). Find the coordinates of its orthocenter.

Let (x_1,y_1)=(0,5),~(x_2,y_2)=(-1,2),~(x_3,y_3)=(3,0). Since the denominators in the expressions for x and y above are the same, we first calculate this “common denominator”:

    \begin{equation*} \begin{split} \textrm{\textbf{denominator}}&=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\\ &=0(2-0)+(-1)(0-5)+3(5-2)\\ &=0(2)+(-1)(-5)+3(3)\\ &=0+5+9\\ &=14 \end{split} \end{equation}

We now calculate the numerator of x:

    \begin{equation*} \begin{split} \textrm{\textbf{x's numerator}}&=x_1x_2(y_2-y_1)+x_2x_3(y_3-y_2)+x_1x_3(y_1-y_3)-(y_2-y_1)(y_3-y_2)(y_1-y_3)\\ &=(0)(-1) (2-5) + (-1)(3)(0-2) + (0)(3)(5-0)-(2-5)(0-2)(5-0)\\ &=(0)(-1)(-3)+(-1)(3)(-2)+(0)(3)(5)-(-3)(-2)(5)\\ &=0+6+0-30\\ &=-24 \end{split} \end{equation}

So, the x-coordinate of the orthocenter is x=\frac{-24}{14}=-\frac{12}{7}.

Next, we calculate the numerator in the expression for y:

    \begin{equation*} \begin{split} \textrm{\textbf{y's numerator}}&=y_1y_2(x_1-x_2)+y_2y_3(x_2-x_3)+y_3y_1(x_3-x_1)-(x_1-x_2)(x_2-x_3)(x_3-x_1)\\ &=(5)(2)(0--1)+(2)(0)(-1-3)+(0)(5)(3-0)-(0--1)(-1-3)(3-0)\\ &=(5)(2)(1)+(2)(0)(-4)+(0)(5)(3)-(1)(-4)(3)\\ &=10+0+0+12\\ &=22 \end{split} \end{equation}

So, the y-coordinate of the orthocenter is \frac{22}{14}=\frac{11}{7}.

Combining the x and y coordinates together, we obtain \Big(-\frac{12}{7},\frac{11}{7}\Big) as the orthocenter.

Let’s solve the same question using the traditional method.

Example 2

Find the orthocenter of \triangle ABC with vertices at A(0,5),~B(-1,2),~C(3,0).

We’ll find the equations of the altitudes through A and C, and then find their point of intersection.

For the altitude through A, its slope is 2 (because it’s the negative reciprocal of the slope of side BC, which has a slope of -\frac{1}{2}). So the equation of the altitude through A can be given by y=2x+b. Using the coordinates of point A(0,5), we see that b=5, and so

(1)   \begin{equation*} y=2x+5 \end{equation*}

For the altitude through C, its slope is -\frac{1}{3} (because it’s the negative reciprocal of the slope of side AB, whose slope is 3). So, the equation of the altitude through C can be given by y=-\frac{1}{3}x+b. Using the coordinates of point C(3,0), we can calculate b:

    \begin{equation*} \begin{split} y&=-\frac{1}{3}x+b\\ 0&=-\frac{1}{3}(3)+b\\ 0&=-1+b\\ 1&=b\\ &\vdots\cdots\vdots\\ y&=-\frac{1}{3}x+1 \end{split} \end{equation}

The altitude through point C has

(2)   \begin{equation*} y=-\frac{1}{3}x+1 \end{equation*}

as its equation. It now remains to solve equations (1) and (2) simultaneously:

    \begin{equation*} \begin{split} y&=2x+5\\ y&=-\frac{1}{3}x+1\\ &\vdots\cdots\vdots\\ 2x+5&=-\frac{1}{3}x+1\\ 3(2x+5)&=3\Big(-\frac{1}{3}x+1\Big)\\ 6x+15&=-x+3\\ 6x+x&=3-15\\ 7x&=-12\\ x&=-\frac{12}{7}\\ &\vdots\cdots\vdots\\ y&=2x+5\\ y&=2\Big(-\frac{12}{7}\Big)+5\\ &=-\frac{24}{7}+5\\ &=-\frac{24}{7}+\frac{5\times 7}{1\times 7}\\ &=-\frac{24}{7}+\frac{35}{7}\\ &=\frac{11}{7} \end{split} \end{equation}

We obtain \Big(-\frac{12}{7},\frac{11}{7}\Big), as before, as expected.

Example 3

Find the coordinates of the orthocenter of \triangle ABC with vertices at A(1,1),~B(3,4),~C(5,-2).

We’ll use the formula now, and then later follow the regular procedure. To this end, let (x_1,y_1)=(1,1),~(x_2,y_2)=(3,4),~(x_3,y_3)=(5,-2). We first find the “common denominator”:

    \begin{equation*} \begin{split} \textrm{\textbf{denominator}}&=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\\ &=1(4--2)+3(-2-1)+5(1-4)\\ &=1(6)+3(-3)+5(-3)\\ &=6-9-15\\ &=-18 \end{split} \end{equation}

Now let’s find the numerator of x expression:

    \begin{equation*} \begin{split} \textrm{\textbf{x'snmrt}}&=x_1x_2(y_2-y_1)+x_2x_3(y_3-y_2)+x_1x_3(y_1-y_3)-(y_2-y_1)(y_3-y_2)(y_1-y_3)\\ &=(1)(3)(4-1)+(3)(5)(-2-4)+(1)(5)(1--2)-(4-1)(-2-4)(1--2)\\ &=3(3)+15(-6)+5(3)-(3)(-6)(3)\\ &=9-90+15+54\\ &=-12 \end{split} \end{equation}

Thus, the x coordinate of the orthocenter is \frac{-12}{-18}=\frac{2}{3}.

Next, we find the numerator of y expression:

    \begin{equation*} \begin{split} \textrm{\textbf{y'snmrt}}&=y_1y_2(x_1-x_2)+y_2y_3(x_2-x_3)+y_3y_1(x_3-x_1)-(x_1-x_2)(x_2-x_3)(x_3-x_1)\\ &=(1)(4)(1-3)+(4)(-2)(3-5)+(-2)(1)(5-1)-(1-3)(3-5)(5-1)\\ &=4(-2)+(-8)(-2)+(-2)(4)-(-2)(-2)(4)\\ &=-8+16-8-16\\ &=-16 \end{split} \end{equation}

Thus, the y coordinate of the orthocenter is \frac{-16}{-18}=\frac{8}{9}.

Combining the x and y coordinates together, we obtain that the orthocenter is located at \Big(\frac{2}{3},\frac{8}{9}\Big).

Example 4

Find the coordinates of the orthocenter of \triangle ABC with vertices at A(1,1),~B(3,4),~C(5,-2).

This time we proceed in the traditional way. This means that we find the equations of any two altitudes. The altitude through point A(1,1) has equation

(3)   \begin{equation*} y=\frac{1}{3}x+\frac{2}{3} \end{equation*}

The altitude through point B(3,4) has equation

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

We now solve equations (3) and (4) simultaneously:

    \begin{equation*} \begin{split} y&=\frac{1}{3}x+\frac{2}{3}\\ y&=\frac{4}{3}x\\ &\vdots\cdots\vdots\\ \frac{1}{3}x+\frac{2}{3}&=\frac{4}{3}x\\ \frac{1}{3}x-\frac{4}{3}x&=-\frac{2}{3}\\ \frac{-3}{3}x&=-\frac{2}{3}\\ -x&=-\frac{2}{3}\\ x&=\frac{2}{3}\\ &\vdots\cdots\vdots\\ y&=\frac{4}{3}x\\ &=\frac{4}{3}\Big(\frac{2}{3}\Big)\\ &=\frac{4\times 2}{3\times 3}\\ &=\frac{8}{9} \end{split} \end{equation}

So we obtain \Big(\frac{2}{3},\frac{8}{9}\Big), as before, as expected.

Example 5

\triangle ABC has vertices at A(-2,3),~B(1,-1),~C(4,5). Find the coordinates of its orthocenter.

As our manner is, we first use the formula. Let (x_1,y_1)=(-2,3),~(x_2,y_2)=(1,-1),~(x_3,y_3)=(4,5). We find the “common denominator” in the expressions for x and y:

    \begin{equation*} \begin{split} \textrm{\textbf{denominator}}&=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\\ &=(-2)(-1-5)+(1)(5-3)+(4)(3--1)\\ &=(-2)(-6)+(1)(2)+(4)(4)\\ &=12+2+16\\ &=30 \end{split} \end{equation}

We now find the numerator in the expression for x:

    \begin{equation*} \begin{split} \textrm{\textbf{x'snmrt}}&=x_1x_2(y_2-y_1)+x_2x_3(y_3-y_2)+x_1x_3(y_1-y_3)-(y_2-y_1)(y_3-y_2)(y_1-y_3)\\ &=(-2)(1)(-1-3)+(1)(4)(5--1)+(4)(-2)(3-5)-(-1-3)(5--1)(3-5)\\ &=(-2)(-4)+(4)(6)+(-8)(-2)-(-4)(6)(-2)\\ &=8+24+16-48\\ &=0 \end{split} \end{equation}

Thus, the x-coordinate of the orthocenter is \frac{0}{30}=0.

Next, we find the numerator in the expression for y:

    \begin{equation*} \begin{split} \textrm{\textbf{y'snmrt}}&=y_1y_2(x_1-x_2)+y_2y_3(x_2-x_3)+y_3y_1(x_3-x_1)-(x_1-x_2)(x_2-x_3)(x_3-x_1)\\ &=(3)(-1)(-2-1)+(-1)(5)(1-4)+(5)(3)(4--2)-(-2-1)(1-4)(4--2)\\ &=(-3)(-3)+(-5)(-3)+(15)(6)-(-3)(-3)(6)\\ &=9+15+90-54\\ &=60 \end{split} \end{equation}

So the y-coordinate of the orthocenter is \frac{60}{30}=2.

Combining the x and y coordinates together, we obtain the point (0,2) as the location of the orthocenter.

Example 6

\triangle ABC has vertices at A(-2,3),~B(1,-1),~C(4,5). Find the coordinates of its orthocenter.

We use the traditional method — always the preferred approach — this time. Let’s find the equations of the altitudes through B and C:

(5)   \begin{equation*} y=-3x+2\quad\implies\textrm{through B} \end{equation*}

(6)   \begin{equation*} y=\frac{3}{4}x+2\quad\implies\textrm{through C} \end{equation*}

And so, at the point of intersection,

    \begin{equation*} \begin{split} -3x+2&=\frac{3}{4}x+2\\ -3x-\frac{3}{4}x&=2-2\\ -\frac{15}{4}x&=0\\ x&=0\\ &\vdots\cdots\vdots\\ &\vdots\vdots\vdots\\ &\vdots\cdots\vdots\\ y&=-3x+2\\ &=-3(0)+2\\ &=2 \end{split} \end{equation}

So we obtain (0,2), as before, as expected.

Much work, little worth

Imagine that you have an appointment in a certain place, but you arrived very much ahead of schedule. So you decided to explore the surrounding area, intentionally whiling away time until your appointment time. Afterward you return to the main area. We’re going to do something similar here. We know that the orthocenter of a right triangle is at the vertex containing 90^{\circ} — that’s our destination which we’ve already arrived at. But we’re going to play around a little bit, and reobtain the result in a different way (Examples 7 to 10).

Example 7

Let A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3) be the vertices of any triangle in which y_1=y_2. Then the x-coordinate of the orthocenter is precisely x_3.

To see this, recall that the x-coordinate of the orthocenter is given by:

x=\frac{x_1x_2(y_2-y_1)+x_2x_3(y_3-y_2)+x_1x_3(y_1-y_3)-(y_2-y_1)(y_3-y_2)(y_1-y_3)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}

And so letting y_1=y_2, the expression reduces to:

    \begin{equation*} \begin{split} x&=\frac{x_2x_3(y_3-y_2)+x_1x_3(y_1-y_3)}{x_1(y_2-y_3)+x_2(y_3-y_1)}\\ &=\frac{x_2x_3(y_3-y_2)+x_1x_3(y_2-y_3)}{x_1(y_2-y_3)+x_2(y_3-y_2)}\quad\textrm{since}~y_1=y_2\\ &=\frac{x_2x_3(y_3-y_2)-x_1x_3(y_3-y_2)}{-x_1(y_3-y_2)+x_2(y_3-y_2)}\\ &=\frac{x_2x_3(y_3-y_2)-x_1x_3(y_3-y_2)}{x_2(y_3-y_2)-x_1(y_3-y_2)}\\ &=\frac{x_3\Big(x_2(y_3-y_2)-x_1(y_3-y_2)\Big)}{x_2(y_3-y_2)-x_1(y_3-y_2)}\\ &=x_3 \end{split} \end{equation}

Example 8

Let A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3) be any triangle in which x_1=x_3. Then the y-coordinate of the orthocenter is just y_2.

To see this, use the fact that the y-coordinate of the orthocenter is given by

y=\frac{y_1y_2(x_1-x_2)+y_2y_3(x_2-x_3)+y_3y_1(x_3-x_1)-(x_1-x_2)(x_2-x_3)(x_3-x_1)}{y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)}

and then let x_1=x_3. The remainder of the proof is similar to that of Example 7, so we omit it.

Example 9

Let A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3) be the vertices of \triangle ABC. Then \triangle ABC is a right triangle if, and only if, there is a rotation (of \triangle ABC) in which two of its sides are parallel to the x and y axes.

The above result is a consequence of the x and y axes being at right angles to each other. Let’s prove one direction (and ask you to prove the other direction in the exercise).

Suppose that there’s a rotation of \triangle ABC in which one side is parallel to the x-axis and another side is parallel to the y-axis. Then the vertex from which these two sides originate is a right angle. Again, this is because: x~\textrm{axis}\perp y~\textrm{axis}.

The point of the above result is that if \triangle ABC with vertices at A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3) is such that \angle A=90^{\circ}, then without loss of generality, we can let x_1=x_3 and y_1=y_2. (If this is not the case with the given triangle, then it can be rotated in such a way that this is the case. And what’s more? Rotation preserves the structure of the triangle.)

Example 10

PROVE that the orthocenter of a right triangle is at the vertex containing angle 90^{\circ}.

We’ve been merry-going-round for this purpose. Let \triangle ABC be a right triangle with vertices at A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3). Suppose that \angle A=90^{\circ}. Following the discussion in Example 9 above, we can suppose, without loss of generality, that x_1=x_3 and y_1=y_2.

Since y_1=y_2, we know that the x-coordinate of the orthocenter is precisely x_3 (cf. Example 7). Moreover, because x_1=x_3, we know that the y-coordinate of the orthocenter is y_2 (cf. Example 8).

Putting these together, we get that the orthocenter is located at (x_3,y_2)=(x_1,y_1), which is point A — the vertex that contains the 90^{\circ} angle.

Have you heard about “much ado about nothing”? You just witnessed one.

Takeaway

The orthocenter formula presented in this post is just to show you what is possible in principle. In practice, we encourage you to always follow the traditional procedure of deriving the equations of two altitudes, then finding their point of intersection to obtain the orthocenter.

Tasks

  1. The orthocenter of \triangle ABC is located at (0,0). Determine an appropriate choice for the coordinates of the vertices A,~B,~C. Is the answer unique?
  2. Find the orthocenter of a triangle whose vertices are located at P(-3,2),~Q(3,0),~(6,2). First use the formula and then use the regular procedure. Verify that both approaches lead to the same result.
  3. PROVE that x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1), and so the denominators in our orthocenter formula are equal.
  4. PROVE that x_1x_2(y_2-y_1)+x_2x_3(y_3-y_2)+x_1x_3(y_1-y_3)-(y_2-y_1)(y_3-y_2)(y_1-y_3)=y_1x_1(x_3-x_2)+y_2x_2(x_1-x_3)+y_3x_3(x_2-x_1)-(y_2-y_1)(y_3-y_2)(y_1-y_3), and so we have a different expression for the numerator of x in the orthocenter formula. [Something similar can be done for the numerator of y.]
  5. Let A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3) be the vertices of \triangle ABC with \angle A=90^{\circ}. PROVE that there is a rotation of \triangle ABC in which sides AB and AC are parallel to the coordinate axes.

Two formulas for the circumcenter

Given a triangle ABC, it is possible to draw a special circle that goes through the three vertices A,B,C. That circle is called the circumscribed circle, and its center is called the circumcenter of the triangle. Knowing the circumcenter is crucial to drawing the circumscribed circle, and the standard way of finding the (coordinates of the) circumcenter consists of two main steps:

  1. find the equations of the right bisectors of any two sides;
  2. solve the equations in step 1 to determine their point of intersection, which will then be the desired circumcenter.

In this post, we present two formulas for finding the coordinates of the circumcenter of a triangle, given the coordinates of the vertices of the triangle. Despite this, the standard way of proceeding outlined above is still the preferred strategy.

Formula 1

Maybe this formula executes as fast as the cars in the popular “Formula 1” sport? You be the judge.

Let A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3) be the vertices of a \triangle ABC. Then the coordinates of its circumcenter are given by:

    \begin{equation*} \begin{split} x&=\frac{x_1^2(y_2-y_3)+x_2^2(y_3-y_1)+x_3^2(y_1-y_2)-(y_1-y_2)(y_2-y_3)(y_3-y_1)}{2\Big(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\Big)}\\ y&=\frac{y_1^2(x_3-x_2)+y_2^2(x_1-x_3)+y_3^2(x_2-x_1)-(x_3-x_2)(x_2-x_1)(x_1-x_3)}{2\Big(y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)\Big)} \end{split} \end{equation}

Beautiful, beautiful thing. Notice that the denominators in both expressions are equal (see the exercises at the end), and so it is enough to calculate just one of them to save time (see Example 2 below).

Example 1

Find the circumcenter of \triangle ABC with vertices at A(0,0),~B(0,3),~C(4,0).

Put A(0,0)=(x_1,y_1),~B(0,3)=(x_2,y_2),~C(4,0)=(x_3,y_3). Then, “Formula 1” above gives

    \begin{equation*} \begin{split} x&=\frac{0^2(3-0)+0^2(0-0)+4^2(0-3)-(0-3)(3-0)(0-0)}{2\Big(0(3-0)+0(0-0)+4(0-3)\Big)}\\ &=\frac{-48}{-24}\\ &=2\\ &\vdots\cdots\vdots\\ y&=\frac{0(4)+9(-4)+0-0}{2\Big(3(-4)\Big)}\\ &=\frac{-36}{-24}\\ &=\frac{3}{2} \end{split} \end{equation}

The circumcenter is at (2,\frac{3}{2}). You can verify this by the standard procedure (or, in this case, by observing that the given triangle is a right triangle, and so the circumcenter is at the midpoint of the hypotenuse).

Example 2

Find the circumcenter of a triangle ABC with vertices at A(-2,1),~B(3,4),~C(6,-5).

Let A(-2,1)=(x_1,y_1),~B(3,4)=(x_2,y_2),~C(6,-5)=(x_3,y_3). We first calculate the “common denominator” in the expressions for x and y:

    \begin{equation*} \begin{split} \textrm{\textbf{denominator}}&=2\Big(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\Big)\\ &=2\Big(-2(9)+3(-6)+6(-3)\Big)\\ &=2(-54)\\ &=-108 \end{split} \end{equation}

This will then appear in the denominators for the x and y coordinates of the circumcenter.

The numerator for the x coordinate is: x_1^2(y_2-y_3)+x_2^2(y_3-y_1)+x_3^2(y_1-y_2)-(y_1-y_2)(y_2-y_3)(y_3-y_1),
where, by substitution, we obtain

    \begin{equation*} \begin{split} &=4(9)+9(-6)+36(-3)-(-3)(9)(-6)\\ &=36-54-108-162\\ &=-288 \end{split} \end{equation}

So the x coordinate of the circumcenter is \frac{-288}{-108}=\frac{8}{3}.

Similarly, the numerator for the y coordinate is:
y_1^2(x_3-x_2)+y_2^2(x_1-x_3)+y_3^2(x_2-x_1)-(x_3-x_2)(x_2-x_1)(x_1-x_3),
where, by substitution, we obtain:

    \begin{equation*} \begin{split} &=1(3)+16(-8)+25(5)-(3)(-8)(5)\\ &=3-128+125+120\\ &=120 \end{split} \end{equation}

So the y coordinate of the circumcenter is \frac{120}{-108}=-\frac{10}{9}. Putting the x and y coordinates together, we obtain \Big(\frac{8}{3},-\frac{10}{9}\Big) as the circumcenter.

We’ll redo this example following the regular procedure. In fact, as we mentioned before, the regular procedure is the best when it comes to numerical problems (as it can be somewhat confusing to keep track of the coordinates in the general formula); the general formula comes in handy when deducing general properties.

Example 3

Find the circumcenter of a triangle ABC with vertices at A(-2,1),~B(3,4),~C(6,-5).

First, a diagram showing the right bisector of AB (dashed line RP) and the right bisector of AC (dashed line NS):

Rendered by QuickLaTeX.com

Let’s find the equations of these right bisectors.

For the bisector RP of side AB, note that since R is the midpoint of AB, then R is located at (\frac{1}{2},\frac{5}{2}). The slope of AB is \frac{3}{5}, so the slope of RP is -\frac{5}{3}. The equation of RP is then:

    \begin{equation*} \begin{split} y&=mx+b\\ y&=-\frac{5}{3}x+b\\ &\vdots\cdots\vdots\\ \frac{5}{2}&=-\frac{5}{3}\Big(\frac{1}{2}\Big)+b\quad\textrm{using point} R(1/2,5/2)\\ \frac{5}{2}&=-\frac{5}{6}+b\\ \frac{5}{2}+\frac{5}{6}&=b\\ \frac{5\times 3}{2\times 3}+\frac{5}{6}&=b\quad\textrm{common denominator}\\ \frac{15}{6}+\frac{5}{6}&=b\\ \frac{20}{6}&=b\\ \frac{10}{3}&=b\quad\textrm{reduce to lowest terms}\\ &\vdots\cdots\vdots\\ \textrm{So,}~y&=-\frac{5}{3}x+\frac{10}{3} \end{split} \end{equation}

Next, we find the equation of the bisector NS of side AC. Note that since N is the midpoint of AC, N is located at (2,-2). The slope of AC is -\frac{3}{4}, so the slope of NS is \frac{4}{3}. Thus, we can write the equation of NS in the form y=\frac{4}{3}x+b, where b can be determined as shown below:

    \begin{equation*} \begin{split} y&=\frac{4}{3}x+b\\ -2&=\frac{4}{3}(2)+b\quad\textrm{using the point}~N(2,-2)\\ -2&=\frac{8}{3}+b\\ -2-\frac{8}{3}&=b\\ \frac{-2\times 3}{1\times 3}-\frac{8}{3}&=b\quad\textrm{common denominator}\\ -\frac{6}{3}-\frac{8}{3}&=b\\ -\frac{14}{3}&=b\\ \vdots\cdots\vdots\\ \textrm{So,}~y&=\frac{4}{3}x-\frac{14}{3} \end{split} \end{equation}

We now solve the two equations simultaneously using substitution.

    \begin{equation*} \begin{split} y&=-\frac{5}{3}x+\frac{10}{3}\\ y&=\frac{4}{3}x-\frac{14}{3}\\ &\vdots\cdots\vdots\\ -\frac{5}{3}x+\frac{10}{3}&=\frac{4}{3}x-\frac{14}{3}\\ -\frac{5}{3}x-\frac{4}{3}x&=-\frac{14}{3}-\frac{10}{3}\\ -\frac{9}{3}x&=-\frac{24}{3}\\ -3x&=-8\\ x&=\frac{8}{3}\\ &\vdots\cdots\vdots\\ y&=\frac{4}{3}x-\frac{14}{3}\\ \implies y&=\frac{4}{3}\Big(\frac{8}{3}\Big)-\frac{14}{3}\\ &=\frac{32}{9}-\frac{14}{3}\\ &=\frac{32}{9}-\frac{14\times 3}{3\times 3}\quad\textrm{common denominator}\\ &=\frac{32}{9}-\frac{42}{9}\\ y&=-\frac{10}{9} \end{split} \end{equation}

So we obtain that the circumcenter is (\frac{8}{3},-\frac{10}{9}), as before.

One of the advantages of having a general formula is that it enables us to see what happens in some special cases.

Example 4

PROVE that if a triangle has two vertices with equal y-coordinates, then the x-coordinate of the circumcenter is the average of the x-coordinates of those two vertices.

To see this, let y_1=y_2 in our formula, then the expression for x reduces to:

    \begin{equation*} \begin{split} x&=\frac{x_1^2(y_2-y_3)+x_2^2(y_3-y_1)}{2\Big(x_1(y_2-y_3)+x_2(y_3-y_1)\Big)}\\ &=\frac{x_1^2(y_2-y_3)+x_2^2(y_3-y_2)}{2\Big(x_1(y_2-y_3)+x_2(y_3-y_2)\Big)}\quad\textrm{using}~y_1=y_2\\ &=\frac{x_1^2(y_2-y_3)-x_2^2(y_2-y_3)}{2\Big(x_1(y_2-y_3)-x_2(y_2-y_3)\Big)}\\ &=\frac{(x_1^2-x_2^2)(y_2-y_3)}{2(x_1-x_2)(y_2-y_3)}\\ &=\frac{(x_1+x_2)(x_1-x_2)(y_2-y_3)}{2(x_1-x_2)(y_2-y_3)}\\ &=\frac{x_1+x_2}{2} \end{split} \end{equation}

Example 5

PROVE that if a triangle has two vertices with equal x-coordinates, then the y-coordinate of the circumcenter is the average of the y-coordinates of those two vertices.

If we let x_1=x_3 in our formula, then the expression for y reduces to:

    \begin{equation*} \begin{split} y&=\frac{y_1^2(x_3-x_2)+y_3^2(x_2-x_1)}{2\Big(y_1(x_3-x_2)+y_3(x_2-x_1)\Big)}\\ &=\frac{y_1^2(x_3-x_2)+y_3^2(x_2-x_3)}{2\Big(y_1(x_3-x_2)+y_3(x_2-x_3)\Big)}\quad\textrm{using}~x_1=x_3\\ &=\frac{y_1^2(x_3-x_2)-y_3^2(x_3-x_2)}{2\Big(y_1(x_3-x_2)-y_3(x_3-x_2)\Big)}\\ &=\frac{(y_1^2-y_3^2)(x_3-x_2)}{2(y_1-y_3)((x_3-x_2))}\\ &=\frac{(y_1+y_3)(y_1-y_3)(x_3-x_2)}{2(y_1-y_3)(x_3-x_2)}\\ &=\frac{y_1+y_3}{2} \end{split} \end{equation}

Since every right triangle can be “oriented” in such a way that its legs are parallel to the x and y axes (making the vertex containing 90^{\circ} to have the same y coordinate with one leg and the same x coordinate with the other leg), we see why the circumcenter of a right triangle is at the midpoint of its hypotenuse side. Alternatively, here is a direct proof using “Formula 1” above.

Notation: Preparation for “Formula 2”

Consider \triangle ABC shown below:

Rendered by QuickLaTeX.com

We’ll denote the slope of side AB by m_{c} (to mimic the regular notation for side AB); similarly, the slope of side AC will be denoted by m_{b}. Don’t confuse these with the coordinates of the vertices.

Formula 2

Let A(a,b),~B(c,d),~C(e,f) be the vertices of a triangle ABC. Then the coordinates of the circumcenter are given by:

    \begin{equation*} \begin{split} x&=\frac{(a+c)m_b-(a+e)m_c+(d-f)m_bm_c}{2(m_b-m_c)}\\ y&=\frac{(b+f)m_b-(b+d)m_c+(e-c)}{2(m_b-m_c)} \end{split} \end{equation}

Let’s consider some examples to illustrate how simple it is to use in practice, so long as the “manual” below is followed:

  1. choose any vertex to start with (call this the “home vertex” — in the diagram above, it is vertex A);
  2. find the slopes of side AB (denoted by m_c) and side AC (denoted by m_b) — the two sides that originate from the “home vertex”;
  3. find the numerator of the x-coordinate:
    • add the x-coordinate of the “home vertex” to the x-coordinate of the vertex on the bottom left, and multiply the sum by the slope of the other side (m_b)
    • add the x-coordinate of the “home vertex” to the x-coordinate of the vertex on the bottom right, and multiply the sum by the slope of the other side (m_c)
    • subtract the y-coordinate of the vertex on the bottom right from the y-coordinate of the vertex on the bottom left, then multiply the difference by m_b\times m_c
    • combine the three results in the order: +,-,+
  4. find the numerator of the y-coordinate:
    • add the y-coordinate of the “home vertex” to the y-coordinate of the vertex on the bottom right, and multiply the sum by the slope of the same side
    • add the y-coordinate of the “home vertex” to the y-coordinate of the vertex on the bottom left, and multiply the sum by the slope of the same side
    • subtract the x-coordinate of the vertex on the bottom right from the x-coordinate of the vertex on the bottom left
    • combine the three results in the order +,-,+
  5. find the denominators of the x and y coordinates (both are 2(m_b-m_c), which is twice the difference of the slopes subtracted in that order);
  6. carry out the divisions;
  7. DONE!!!

Too many steps? Probably confusing? It’s easier than you think.

Example 6

\triangle ABC has vertices at A(3,2),~B(1,4),~C(5,4). Find the coordinates of its circumcenter.

Let A(3,2) be our “home vertex”. For convenience, we’ve drawn the triangle as follows:

Rendered by QuickLaTeX.com

Now, let’s calculate the x-coordinate of the circumcenter:

    \begin{equation*} \begin{split} x&=\frac{(a+c)m_b-(a+e)m_c+(d-f)m_bm_c}{2(m_b-m_c)}\\ &=\frac{(3+1)(1)-(3+5)(-1)+(4-4)(1)(-1)}{2(1--1)}\\ &=\frac{4+8+0}{4}\\ &=\frac{12}{4}\\ &=3 \end{split} \end{equation}

Next, we find the y-coordinate of the circumcenter:

    \begin{equation*} \begin{split} y&=\frac{(b+f)m_b-(b+d)m_c+(e-c)}{2(m_b-m_c)}\\ &=\frac{(2+4)(1)-(2+4)(-1)+(5-1)}{2(1--1)}\\ &=\frac{6+6+4}{4}\\ &=\frac{16}{4}\\ &=4 \end{split} \end{equation}

So, the circumcenter is located at (3,4). Note that if the diagram was drawn accurately, then \triangle ABC is a right triangle with BC as the hypotenuse. The midpoint of BC is (3,4), the same as the circumcenter obtained. This is an instance of the well-known result we mentioned before.

To show that the answer is independent of which vertex was chosen as the “home vertex”, we’ll solve the example above again, using a different “home vertex”.

Example 7

\triangle ABC has vertices at A(3,2),~B(1,4),~C(5,4). Find the coordinates of its circumcenter.

Let’s choose B(1,4) as the “home vertex”. As usual, we draw the diagram for convenience:

Rendered by QuickLaTeX.com

Ideally we should have m_a instead of m_b in the above diagram, but to avoid confusion, we’ve retained the old notation.

The x-coordinate of the circumcenter is:

    \begin{equation*} \begin{split} x&=\frac{(a+c)m_b-(a+e)m_c+(d-f)m_bm_c}{2(m_b-m_c)}\\ &=\frac{(1+3)(0)-(1+5)(-1)+(2-4)(0)(-1)}{2(0--1)}\\ &=\frac{0+6+0}{2}\\ &=\frac{6}{2}\\ &=3 \end{split} \end{equation}

Then we find the y-coordinate of the circumcenter:

    \begin{equation*} \begin{split} y&=\frac{(b+f)m_b-(b+d)m_c+(e-c)}{2(m_b-m_c)}\\ &=\frac{(4+4)(0)-(4+2)(-1)+(4-2)}{2(0--1)}\\ &=\frac{0+6+2}{2}\\ &=\frac{8}{2}\\ &=4 \end{split} \end{equation}

So we obtain (3,4) as before. That should do.

Example 8

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

This time around we proceed in the traditional way. First, a diagram:

Rendered by QuickLaTeX.com

This diagram shows where the circumcenter is (Q). Nevertheless, we still carry out the calculations.

The midpoint of AC is N(0,-1) and the slope of AC is 1. Thus the slope of the right bisector NR of AC is -1, and its equation is:

(1)   \begin{equation*} y=-x-1 \end{equation*}

The midpoint of BC is J(1,0.5) and the slope of BC is \frac{1}{2}. Thus the slope of the right bisector JK of BC is -2, and its equation is:

(2)   \begin{equation*} y=-2x+\frac{5}{2} \end{equation*}

We now have to solve equation(1) and equation(2) simultaneously:

    \begin{equation*} \begin{split} y&=\boxed{-x-1}\\ y&=\boxed{-2x+\frac{5}{2}}\\ -x-1&=-2x+\frac{5}{2}\\ -x+2x&=\frac{5}{2}+1\\ x&=\frac{5}{2}+\frac{2}{2}\\ x&=\frac{7}{2}\\ &\vdots\cdots\vdots\cdots\vdots\cdots\vdots\\ y&=-x-1\\ y&=-\frac{7}{2}-1\\ y&=-\frac{7}{2}-\frac{2}{2}\\ y&=-\frac{9}{2} \end{split} \end{equation}

Therefore, the circumcenter is located at the point Q(\frac{7}{2},-\frac{9}{2}), as seen in the diagram.

Let’s solve the same question using our formula.

Example 9

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

We draw a convenient diagram to begin with:

Rendered by QuickLaTeX.com

Again we should have used m_a instead of m_b, but we want to stick to the old notation to avoid confusion.

The x-coordinate of the circumcenter is:

    \begin{equation*} \begin{split} x&=\frac{(a+c)m_b-(a+e)m_c+(d-f)m_bm_c}{2(m_b-m_c)}\\ &=\frac{(0+-2)(1/2)-(0+2)(3/2)+(-2-2)(1/2)(3/2)}{2(1/2-3/2)}\\ &=\frac{-2(1/2)-2(3/2)-4(3/4)}{2(-2/2)}\\ &=\frac{-1-3-3}{2(-1)}\\ &=\frac{-7}{-2}\\ &=\frac{7}{2} \end{split} \end{equation}

The y-coordinate of the circumcenter is:

    \begin{equation*} \begin{split} y&=\frac{(b+f)m_b-(b+d)m_c+(e-c)}{2(m_b-m_c)}\\ &=\frac{(0+1)(1/2)-(0+-3)(3/2)+(2--2)}{2(1/2-3/2)}\\ &=\frac{1/2+9/2+4}{2(-1)}\\ &=\frac{9}{-2}\\ &=-\frac{9}{2} \end{split} \end{equation}

We obtain the same result (\frac{7}{2},-\frac{9}{2}), as expected.

Example 10

Find the circumcenter of the right triangle ABC with vertices A(-h,0),~B(h,0),~C(0,h), where h\neq 0.

We begin with a convenient diagram:

Rendered by QuickLaTeX.com

The x-coordinate of the circumcenter is:

    \begin{equation*} \begin{split} x&=\frac{(a+c)m_b-(a+e)m_c+(d-f)m_bm_c}{2(m_b-m_c)}\\ &=\frac{(0+-h)(-1)-(0+h)(1)+(0-0)(-1)(1)}{2(-1-1)}\\ &=\frac{h-h+0}{-4}\\ &=\frac{0}{-4}\\ &=0 \end{split} \end{equation}

The y-coordinate of the circumcenter is:

    \begin{equation*} \begin{split} y&=\frac{(b+f)m_b-(b+d)m_c+(e-c)}{2(m_b-m_c)}\\ &=\frac{(h+0)(-1)-(h+0)(1)+(h--h)}{2(-1-1)}\\ &=\frac{-h-h+2h}{-4}\\ &=\frac{0}{-4}\\ &=0 \end{split} \end{equation}

So the circumcenter is at (0,0).

Takeaway

If you’ve calculated the area of a triangle or the volume of a cylinder or the amount of a compound interest, and so on, you’ll agree that there are quite a lot of useful formulas in math — in particular, in geometry.

There are certain occasions where the use of those formulas is inevitable; as such, working from “first principles” (or “re-inventing the wheel”) is inadvisable. On the other hand, there are times when working from “first principles” is the better strategy — not because no formula exists, but because it assists in deepening our understanding of what the formula consists. Knowing which approach to use when you have a choice is a matter of preference, and can be mastered through practice.

Think about this for an analogy. Suppose you’re typing and you need to type or copy paste a particular word which you’ve already typed before, one more time. Will you choose to re-type the word or will you choose to copy paste? It’s a matter of preference definitely. Now suppose that you need to type or copy paste a particular paragraph which has already appeared before, one more time. Will you choose to re-type the paragraph or will you just copy paste?

Tasks

  1. Find the circumcenter of an equilateral triangle whose vertices are A(-a,0),~B(a,0),~C(0,a\sqrt{3});\quad a\neq 0.
  2. Find the circumcenter of the isosceles triangle whose vertices are located at P(-a,0),~Q(a,0),~R(0,a);\quad a\neq 0.
  3. Let (x_1,y_1),~(x_2,y_2),~(x_3,y_3) be the vertices of a triangle. If

        \begin{equation*} \begin{split} (x_3-x_1)^2+(y_3-y_1)^2&= (x_2-x_1)^2+(y_2-y_1)^2\\ &+(x_3-x_2)^2+(y_3-y_2)^2,\end{split} \end{equation}

    PROVE that:

        \[\frac{y_3-y_2}{x_3-x_2}\times\frac{y_2-y_1}{x_2-x_1}=-1.\]

    [The converse is also true. Note that the result holds whether or not we’re dealing with the vertices of a triangle. However, when applied to triangles, it tells us that we can either use lengths or slopes (not both) to verify that a triangle contains 90^{\circ}.]

  4. PROVE that:

        \begin{equation*} \begin{split} x_1^2(y_2-y_3)+x_2^2(y_3-y_1)+x_3^2(y_1-y_2)&=y_1(x_3^2-x_2^2)+\\ &y_2(x_1^2-x_3^2)+y_3(x_2^2-x_1^2) \end{split} \end{equation}

    [It follows that we have two expressions for the numerator of the x-coordinate of our circumcenter formula 1. Same thing applies to the y-coordinate.]

  5. PROVE that:

        \begin{equation*} \begin{split} x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)&=y_1(x_3-x_2)+\\ &y_2(x_1-x_3)+y_3(x_2-x_1) \end{split} \end{equation}

    [It follows that we have two expressions for the denominator of the x-coordinate of our circumcenter formula 1. Same thing applies to the y-coordinate.]