The auto center – High School Geometry

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:

find the equations of any two altitudes;
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

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?
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.
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.
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 $y$ .]
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.