Triangle centres and pairs of lines

Area, and three families of lines

A triangle with corners $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ has area $\dfrac{1}{2}|D|$. Here $D$ is the sum of three products: $x_1(y_2 - y_3)$, $x_2(y_3 - y_1)$ and $x_3(y_1 - y_2)$.

Three families of lines inside a triangle give its centres.

• A median joins a corner to the middle point of the opposite side.
• An altitude goes from a corner at right angles to the opposite side. Its slope is the negative reciprocal of the slope of that side.
• A perpendicular bisector goes through the middle point of a side at right angles to it.

The four centres

• Centroid $G$. Where the medians meet. It is the average of the corners, $\left(\dfrac{x_1 + x_2 + x_3}{3},\ \dfrac{y_1 + y_2 + y_3}{3}\right)$.
• Circumcentre $O$. Where the perpendicular bisectors meet. It is equally far, the distance $R$, from all three corners.
• Orthocentre $H$. Where the altitudes meet.
• Incentre $I$. Where the angle bisectors meet. If the sides opposite $A$, $B$ and $C$ have lengths $a$, $b$ and $c$, then $I = \dfrac{aA + bB + cC}{a + b + c}$. The inradius is $r = \dfrac{\text{area}}{s}$, where $s$ is half the perimeter.

Example 1. Take $A(4, 6)$, $B(-3, 5)$ and $C(6, 2)$.

• Area. The three products in the corner formula are $4 \times 3 = 12$, $(-3)(-4) = 12$ and $6 \times 1 = 6$. They add up to 30, so the area is $\dfrac{30}{2} = 15$.
• Centroid. $G = \left(\dfrac{4 - 3 + 6}{3},\ \dfrac{6 + 5 + 2}{3}\right) = \left(\dfrac{7}{3},\ \dfrac{13}{3}\right)$.
• Circumcentre. The perpendicular bisector of $AB$ is $7x + y = 9$, and that of $AC$ is $x - 2y = -3$. They meet at $O(1, 2)$, and $R = OA = 5$.
• Orthocentre. $BC$ has slope $-\dfrac{1}{3}$, so the altitude from $A$ has slope 3: $y = 3x - 6$. $AC$ has slope $-2$, so the altitude from $B$ is $2y = x + 13$. They meet at $H(5, 9)$.

The Euler line

Look at the three points $O(1, 2)$, $G\left(\dfrac{7}{3}, \dfrac{13}{3}\right)$ and $H(5, 9)$. They lie on one line, the Euler line, and $G$ is one third of the way from $O$ to $H$:

Here $3G - 2O = (7 - 2,\ 13 - 4) = (5, 9)$. This is the quickest way to find $H$, because $G$ and $O$ are easy. The Euler line is $7x - 4y + 1 = 0$, through $O$ and $H$.

Three special cases are worth knowing. In an equilateral triangle all four centres coincide. In a right-angled triangle $H$ is the corner with the right angle, and $O$ is the middle point of the hypotenuse. In an obtuse triangle $H$ and $O$ lie outside the triangle.

Example 2 (the incentre). Take $A(0, 0)$, $B(14, 0)$ and $C(5, 12)$. The sides are $a = BC = 15$, $b = CA = 13$ and $c = AB = 14$.

The weighted corners add up to $15(0, 0) + 13(14, 0) + 14(5, 12) = (252, 168)$. Divide by $a + b + c = 42$:

The area is $\dfrac{1}{2} \times 14 \times 12 = 84$ and $s = 21$, so $r = \dfrac{84}{21} = 4$. This agrees with the $y$-coordinate of $I$, since $AB$ lies along the $x$-axis.

Pairs of lines through the origin

An equation in which every term has degree 2, $ax^2 + 2hxy + by^2 = 0$, gives two lines through the origin. Divide by $x^2$ and put $m = \dfrac{y}{x}$:

The two roots $m_1$ and $m_2$ are the slopes of the lines.

• $m_1 + m_2 = -\dfrac{2h}{b}$ and $m_1m_2 = \dfrac{a}{b}$.
• The lines are real and different when $h^2 > ab$, they coincide when $h^2 = ab$, and there is no real line when $h^2 < ab$.
• The acute angle between them has $\tan\theta = \dfrac{2\sqrt{h^2 - ab}}{|a + b|}$.
• The lines are perpendicular when $a + b = 0$.
• The bisectors of the pair satisfy $\dfrac{x^2 - y^2}{a - b} = \dfrac{xy}{h}$.

Example 3. For $2x^2 - 7xy + 3y^2 = 0$ we have $a = 2$, $h = -\dfrac{7}{2}$ and $b = 3$. It factorises as $(2x - y)(x - 3y) = 0$, so the lines are $y = 2x$ and $y = \dfrac{x}{3}$. The formula gives $\tan\theta = \dfrac{2\sqrt{49/4 - 6}}{5} = 1$, an angle of 45 degrees. Check with the slopes: $\dfrac{2 - 1/3}{1 + 2/3} = 1$.

A general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is a pair of lines when

Example 4. For which $k$ is $2x^2 + 5xy + 2y^2 + 3x + 3y + k = 0$ a pair of lines?

Here $a = b = 2$, $h = \dfrac{5}{2}$, $g = f = \dfrac{3}{2}$ and $c = k$. The condition becomes $4k + \dfrac{45}{4} - \dfrac{9}{2} - \dfrac{9}{2} - \dfrac{25k}{4} = 0$, so $\dfrac{9 - 9k}{4} = 0$ and $k = 1$. Then the equation is $(2x + y + 1)(x + 2y + 1) = 0$, and the two lines cross at $\left(-\dfrac{1}{3}, -\dfrac{1}{3}\right)$.

Practise with new numbers

Try it in Python

The program finds the four centres of a triangle, checks that the altitudes are perpendicular to the sides, and shows that $GH$ is twice $OG$ for random triangles.

Common mistakes

• Using the slope of the side. An altitude has the negative reciprocal of that slope.
• Weighting the incentre by the wrong sides. The weight of corner $A$ is $a$, the side opposite it.
• Expecting $H$ inside every triangle. In an obtuse triangle the orthocentre lies outside.
• Reading $2h$ as $h$. In $ax^2 + 2hxy + by^2$, an $xy$ term of $5xy$ means $h = 2.5$.

Remember these

• $G$ is the average of the corners, and $H = 3G - 2O$, so $O$, $G$, $H$ lie on one line with $GH = 2\,OG$.
• $O$ is equally far from the corners, $H$ is where the altitudes meet, and $I = \dfrac{aA + bB + cC}{a + b + c}$.
• For $ax^2 + 2hxy + by^2 = 0$, $\tan\theta = \dfrac{2\sqrt{h^2 - ab}}{|a + b|}$, and the lines are perpendicular when $a + b = 0$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the orthocentre and the circumcentre of the triangle with corners $(0, 0)$, $(8, 0)$ and $(0, 6)$.

2. For the same triangle, find the incentre and the inradius.

3. Find the two lines given by $x^2 - 5xy + 6y^2 = 0$, and the tangent of the angle between them.

4. For which $k$ are the lines of $kx^2 + 6xy - y^2 = 0$ at right angles?