Concurrencies — Mathslive.ie

Section 1 of 3

Centroid — centre of gravity

A median joins a vertex to the midpoint of the opposite side.

Every triangle has three medians — one from each corner.

The three medians always cross at a single point. That point is the centroid.

It's the centre of gravity — balance the triangle on a pin there and it sits level.

The good news: you never have to find the medians by hand. The centroid has a one-line formula.

Must learn

1.Centroid $= \left(\dfrac{x_1+x_2+x_3}{3},\ \dfrac{y_1+y_2+y_3}{3}\right)$

2.Add the three $x$'s, divide by $3$. Same for the $y$'s.

(i) Worked example

Find the centroid of $(3,-1)$, $(4,7)$ and $(2,9)$.

$x$-coordinate: $\dfrac{3+4+2}{3} = \dfrac{9}{3} = 3$

$y$-coordinate: $\dfrac{-1+7+9}{3} = \dfrac{15}{3} = 5$

Centroid $= (3,\ 5)$

Three additions and two divisions. No simultaneous equations — that's what makes the centroid the easy one.

YOU TRY · 1

Find the centroid of $(1,2)$, $(4,6)$ and $(7,1)$.

Average the $x$'s, then average the $y$'s.

$x: \dfrac{1+4+7}{3} = \dfrac{12}{3} = 4$

$y: \dfrac{2+6+1}{3} = \dfrac{9}{3} = 3$

$(4,\ 3)$

YOU TRY · 2

Find the centroid of $(-2,5)$, $(0,-3)$ and $(8,4)$.

Mind the signs.

$x: \dfrac{-2+0+8}{3} = \dfrac{6}{3} = 2$

$y: \dfrac{5-3+4}{3} = \dfrac{6}{3} = 2$

$(2,\ 2)$

YOU TRY · 3

Two vertices of a triangle are $(1,1)$ and $(4,7)$. The centroid is $(3,4)$. Find the third vertex.

Work backwards. Let the third vertex be $(x,y)$ and use the formula.

$x: \dfrac{1+4+x}{3} = 3 \;\Rightarrow\; 5+x = 9 \;\Rightarrow\; x = 4$

$y: \dfrac{1+7+y}{3} = 4 \;\Rightarrow\; 8+y = 12 \;\Rightarrow\; y = 4$

$(4,\ 4)$

Section 2 of 3

Circumcentre — equidistant from each vertex

The circumcentre is the one point that is the same distance from all three vertices.

Because it's equidistant, a circle drawn from it passes through $A$, $B$ and $C$ — the circumscribed circle.

How do we find it? Use the perpendicular bisector of a side.

A perpendicular bisector cuts a side in half at a right angle. Every point on it is the same distance from the two ends of that side.

Do this for two sides. Where the two bisectors cross is the circumcentre.

Method

1.Midpoint of a side.

2.Slope of that side → flip it and change the sign for the perpendicular slope.

3.Line through the midpoint: $y - y_1 = m(x - x_1)$.

4.Repeat for a second side.

5.Solve the two lines simultaneously.

(i) Worked example

Find the circumcentre of $(5,-3)$, $(3,-1)$ and $(-1,5)$.

Side 1 — $(3,-1)$ and $(-1,5)$

Midpoint: $\left(\dfrac{3+(-1)}{2},\ \dfrac{-1+5}{2}\right) = (1,\ 2)$

Slope of the side: $m = \dfrac{5+1}{-1-3} = \dfrac{6}{-4} = -\dfrac{3}{2}$

Perpendicular slope: $m = \dfrac{2}{3}$

Line through $(1,2)$: $y - 2 = \dfrac{2}{3}(x - 1)$

$3y - 6 = 2x - 2$

$2x - 3y = -4$

Side 2 — $(5,-3)$ and $(3,-1)$

Midpoint: $\left(\dfrac{5+3}{2},\ \dfrac{-3+(-1)}{2}\right) = (4,\ -2)$

Slope of the side: $m = \dfrac{-1+3}{3-5} = \dfrac{2}{-2} = -1$

Perpendicular slope: $m = 1$

Line through $(4,-2)$: $y + 2 = 1(x - 4)$

$x - y = 6$

Solve the two together

$2x - 3y = -4 \qquad \text{and} \qquad x - y = 6$

$\times 3$ the second: $3x - 3y = 18$

Subtract from the first: $(2x - 3y) - (3x - 3y) = -4 - 18$

$-x = -22 \quad\Rightarrow\quad x = 22$

Back into $x - y = 6$: $22 - y = 6 \quad\Rightarrow\quad y = 16$

Circumcentre $= (22,\ 16)$

Don't panic that it lands miles outside the triangle — this triangle is obtuse, so its circumcentre sits well outside it. Check it if you like: $(22,16)$ really is the same distance from all three corners.

YOU TRY · 4

Find the perpendicular bisector of the segment joining $(2,1)$ and $(6,5)$.

Midpoint, then perpendicular slope, then the line.

Midpoint $= (4,\ 3)$

Slope $= \dfrac{5-1}{6-2} = 1 \;\Rightarrow\;$ perp slope $= -1$

$y - 3 = -1(x - 4)$

$x + y = 7$

YOU TRY · 5

Find the circumcentre of $(1,1)$, $(5,1)$ and $(1,5)$.

Two of the sides are flat and upright — their perpendicular bisectors are nice and simple.

Side $(1,1)(5,1)$: midpoint $(3,1)$, side is horizontal → bisector is the vertical line $x = 3$.

Side $(1,1)(1,5)$: midpoint $(1,3)$, side is vertical → bisector is the horizontal line $y = 3$.

Circumcentre $= (3,\ 3)$

$(3,\ 3)$

YOU TRY · 6

Find the circumcentre of $(0,0)$, $(8,0)$ and $(4,4)$.

Use the base for the easy bisector, then one slanted side.

Side $(0,0)(8,0)$: midpoint $(4,0)$, horizontal → bisector $x = 4$.

Side $(0,0)(4,4)$: midpoint $(2,2)$, slope $1 \Rightarrow$ perp $-1$: $x + y = 4$.

At $x = 4$: $4 + y = 4 \Rightarrow y = 0$.

Circumcentre $= (4,\ 0)$

$(4,\ 0)$

Section 3 of 3

Orthocentre — where the altitudes meet

An altitude runs from a vertex, straight in to the opposite side, at a right angle.

It's the shortest path from the corner to the far side — perpendicular distance.

All three altitudes meet at one point: the orthocentre.

Same machinery as the circumcentre, but watch the one difference:

Circumcentre vs Orthocentre

○Circumcentre: perpendicular to a side, through that side's midpoint.

●Orthocentre: perpendicular to a side, through the opposite vertex.

Both use a perpendicular slope. The circumcentre goes through the midpoint; the orthocentre goes through the vertex.

Method

1.Slope of a side.

2.Perpendicular slope.

3.Line through the opposite vertex: $y - y_1 = m(x - x_1)$.

4.Repeat for a second side.

5.Solve the two lines simultaneously.

(i) Worked example

Find the orthocentre of $(0,3)$, $(7,4)$ and $(4,-5)$.

Altitude from $(0,3)$

Perpendicular to the opposite side, $(7,4)$ to $(4,-5)$.

Slope of that side: $m = \dfrac{-5-4}{4-7} = \dfrac{-9}{-3} = 3$

Perpendicular slope: $m = -\dfrac{1}{3}$

Through the vertex $(0,3)$: $y - 3 = -\dfrac{1}{3}(x - 0)$

$3y - 9 = -x$

$x + 3y = 9$

Altitude from $(4,-5)$

Perpendicular to the opposite side, $(0,3)$ to $(7,4)$.

Slope of that side: $m = \dfrac{4-3}{7-0} = \dfrac{1}{7}$

Perpendicular slope: $m = -7$

Through the vertex $(4,-5)$: $y + 5 = -7(x - 4)$

$y + 5 = -7x + 28$

$7x + y = 23$

Solve the two together

$x + 3y = 9 \qquad \text{and} \qquad 7x + y = 23$

$\times 3$ the second: $21x + 3y = 69$

Subtract the first: $(21x + 3y) - (x + 3y) = 69 - 9$

$20x = 60 \quad\Rightarrow\quad x = 3$

Back into $x + 3y = 9$: $3 + 3y = 9 \quad\Rightarrow\quad y = 2$

Orthocentre $= (3,\ 2)$

YOU TRY · 7

A triangle has a side from $(1,2)$ to $(5,2)$. The opposite vertex is $(3,7)$. Find the altitude from that vertex.

The side is flat. What's perpendicular to a flat line?

Side slope $= 0$ (horizontal) → altitude is vertical.

Through $(3,7)$.

$x = 3$

YOU TRY · 8

Find the orthocentre of $(0,0)$, $(4,0)$ and $(1,3)$.

Altitude from a vertex ⟂ to the opposite side. Do two of them.

From $(0,0)$ ⟂ side $(4,0)(1,3)$: slope $-1 \Rightarrow$ perp $1$: $y = x$.

From $(4,0)$ ⟂ side $(0,0)(1,3)$: slope $3 \Rightarrow$ perp $-\dfrac{1}{3}$: $x + 3y = 4$.

Sub $y = x$: $4x = 4 \Rightarrow x = 1,\ y = 1$.

Orthocentre $= (1,\ 1)$

$(1,\ 1)$

YOU TRY · 9

Find the orthocentre of $(0,0)$, $(4,0)$ and $(0,3)$.

Look at the angle at $(0,0)$ before you reach for algebra.

From $(4,0)$ ⟂ the vertical side $(0,0)(0,3)$: horizontal line $y = 0$.

From $(0,3)$ ⟂ the flat side $(0,0)(4,0)$: vertical line $x = 0$.

They meet at $(0,0)$ — the right-angle corner.

Orthocentre $= (0,\ 0)$

$(0,\ 0)$ (in a right-angled triangle the orthocentre is the right-angle vertex)

SUM

The three centres in one box

Concurrency toolkit

1.Centroid $= \left(\dfrac{x_1+x_2+x_3}{3},\ \dfrac{y_1+y_2+y_3}{3}\right)$ — average the corners. No equations.

2.Circumcentre — perpendicular bisector of two sides (perp slope, through the MIDPOINT), solve together. Equidistant from all vertices.

3.Orthocentre — altitude from two vertices (perp slope, through the VERTEX), solve together. Where the altitudes meet.

4.Same perpendicular-slope step both times — only the point changes: midpoint vs vertex.

5.Perpendicular slope = flip the fraction and change the sign.

End of lesson

Concurrencies — HL · Mathslive.ie