Centroid and area of a triangle

Two more for­mu­las com­plete the toolkit of this chap­ter. The first gives the cen­troid of a tri­an­gle, the point where its three medi­ans meet, which is also its bal­anc­ing point. The sec­ond gives the area of a tri­an­gle straight from the coor­di­nates of its ver­tices, with no need to find a base or a height. The area for­mula also gives a neat test for whether three points are collinear.

The cen­troid#

A median of a tri­an­gle joins a ver­tex to the mid­point of the oppo­site side. The three medi­ans always meet at one point, the cen­troid $G$, which divides each median in the ratio $2 : 1$ count­ing from the ver­tex.

Cen­troid for­mula. For a tri­an­gle with ver­tices $A(x_1, y_1)$, $B(x_2, y_2)$ and $C(x_3, y_3)$, the cen­troid is

$$\displaystyle G\left(\frac{x_1 + x_2 + x_3}{3}, \; \frac{y_1 + y_2 + y_3}{3}\right)$$

Why. The mid­point of $BC$ is $\displaystyle D\left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right)$. The point divid­ing $AD$ in the ratio $2 : 1$ has x-coor­di­nate

$$\displaystyle \frac{2 \times \frac{x_2 + x_3}{2} + 1 \times x_1}{2 + 1} = \frac{x_1 + x_2 + x_3}{3}$$

and the same for $y$. The answer is sym­met­ric in $A$, $B$, $C$, so the same point lies on all three medi­ans.

Exam­ple 1#

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

$$\displaystyle G = \left(\frac{2 - 4 + 8}{3}, \; \frac{3 + 5 - 2}{3}\right) = (2, 2)$$

Exam­ple 2#

Two ver­tices of a tri­an­gle are $(1, 2)$ and $(3, 5)$, and its cen­troid is $(2, 3)$. Find the third ver­tex.

Let the third ver­tex be $(x, y)$. Then $\displaystyle \frac{1 + 3 + x}{3} = 2$ gives $x = 2$, and $\displaystyle \frac{2 + 5 + y}{3} = 3$ gives $y = 2$. The third ver­tex is $(2, 2)$.

Area of a tri­an­gle#

Area for­mula. The area of the tri­an­gle with ver­tices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$ is

$$\displaystyle \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|$$

The for­mula comes from drop­ping per­pen­dic­u­lars from the three ver­tices to the x-axis: the tri­an­gle's area is then a sum and dif­fer­ence of three trapez­i­ums, and sim­pli­fy­ing gives the expres­sion above. The ver­ti­cal bars mean "take the value with­out its sign", since an area can­not be neg­a­tive.

To use it with­out slips, write the pat­tern care­fully: each $x$ mul­ti­plies the dif­fer­ence of the other two $y$'s, taken in cyclic order ($2 - 3$, then $3 - 1$, then $1 - 2$).

Exam­ple 3#

Find the area of the tri­an­gle with ver­tices $A(1, 1)$, $B(4, 5)$ and $C(8, 2)$.

$$\displaystyle \begin{aligned} \text{Area} &= \frac{1}{2}\left|1(5 - 2) + 4(2 - 1) + 8(1 - 5)\right| \\ &= \frac{1}{2}\left|3 + 4 - 32\right| = \frac{1}{2} \times 25 = 12.5 \text{ square units} \end{aligned}$$

In the les­son on the dis­tance for­mula we showed this tri­an­gle is right-angled at $B$ with $AB = BC = 5$. Its area is there­fore $\displaystyle \frac{1}{2} \times 5 \times 5 = 12.5$, which agrees.

Collinear points#

Three points lie on a straight line exactly when the "tri­an­gle" they form has zero area.

Exam­ple 4#

Show that $(1, 2)$, $(3, 6)$ and $(4, 8)$ are collinear.

$$\displaystyle \frac{1}{2}\left|1(6 - 8) + 3(8 - 2) + 4(2 - 6)\right| = \frac{1}{2}\left|-2 + 18 - 16\right| = 0$$

The area is zero, so the points are collinear. (We found the same result with dis­tances in the pre­vi­ous les­son; this way is shorter.)

Exam­ple 5#

Find $k$ if the points $(k, 3)$, $(2, 5)$ and $(4, 9)$ are collinear.

$$k(5 - 9) + 2(9 - 3) + 4(3 - 5) = 0 \implies -4k + 12 - 8 = 0 \implies k = 1$$

Area of a quadri­lat­eral#

Split the quadri­lat­eral into two tri­an­gles by a diag­o­nal and add their areas. Take the ver­tices in order round the fig­ure.

Exam­ple 6#

Find the area of the quadri­lat­eral with ver­tices $A(0, 0)$, $B(6, 0)$, $C(8, 5)$ and $D(1, 4)$, taken in order.

The diag­o­nal $AC$ splits it into tri­an­gles $ABC$ and $ACD$.

$$\displaystyle \text{ar}(ABC) = \frac{1}{2}\left|0(0 - 5) + 6(5 - 0) + 8(0 - 0)\right| = \frac{1}{2} \times 30 = 15$$

$$\displaystyle \text{ar}(ACD) = \frac{1}{2}\left|0(5 - 4) + 8(4 - 0) + 1(0 - 5)\right| = \frac{1}{2} \times 27 = 13.5$$

The area of the quadri­lat­eral is $15 + 13.5 = 28.5$ square units.

Prac­tice#

1. Find the cen­troid of the tri­an­gle with ver­tices $(1, 4)$, $(-3, 2)$ and $(5, -3)$.
2. The cen­troid of a tri­an­gle is the ori­gin, and two ver­tices are $(3, -2)$ and $(-5, 4)$. Find the third ver­tex.
3. Find the area of the tri­an­gle with ver­tices $(0, 0)$, $(4, 0)$ and $(0, 6)$.
4. Find the area of the tri­an­gle with ver­tices $(2, 3)$, $(-1, 0)$ and $(2, -4)$.
5. Find the area of the tri­an­gle with ver­tices $(-5, -1)$, $(3, -5)$ and $(5, 2)$.
6. Are the points $(-2, -3)$, $(1, 0)$ and $(5, 4)$ collinear?
7. Find $k$ if $(2, -1)$, $(k, 3)$ and $(5, 5)$ are collinear.
8. Find the area of the quadri­lat­eral with ver­tices $(1, 1)$, $(5, 1)$, $(6, 4)$ and $(2, 5)$, taken in order.
9. A tri­an­gle has ver­tices $(0, 0)$, $(8, 0)$ and $(0, 6)$. Find its area and the area of the tri­an­gle formed by the mid­points of its sides.

Answers#