Area of a Triangle — Mathslive.ie

Section 1 of 3

The formula

The area of a triangle straight from its coordinates:

$\text{Area} = \dfrac{1}{2}\,\big|\,x_1 y_2 - x_2 y_1\,\big|$

The bars mean absolute value — an area is never negative.

The catch

1.One corner of the triangle must be at the origin $(0,0)$.

2.If none is, slide the whole triangle until a corner lands there.

(i) Worked example — $\triangle PQR$

Find the area of $\triangle PQR$ where $P(-1,3)$, $Q(2,-5)$ and $R(4,7)$.

No corner is at the origin, so slide $P$ to $(0,0)$. Whatever you do to $P$, do to all three.

$P(-1,3)$ to $(0,0)$ needs $+1,\ -3$. Apply to each:

$P(-1,3) \to (0,0)$

$Q(2,-5) \to (3,-8)$

$R(4,7) \to (5,4)$

Take the two non-origin corners: $(x_1,y_1) = (3,-8)$, $(x_2,y_2) = (5,4)$.

$\text{Area} = \dfrac{1}{2}\,\big|\,3(4) - 5(-8)\,\big|$

$= \dfrac{1}{2}\,\big|\,12 + 40\,\big| = \dfrac{1}{2}(52)$

$= 26$ square units

It doesn't matter which corner you send to the origin — slide $Q$ instead and you still land on $26$.

Slide $Q$ to $(0,0)$ instead ($-2,\ +5$): $P \to (-3,8)$, $R \to (2,12)$.

$\dfrac{1}{2}\,\big|\,(-3)(12) - 2(8)\,\big| = \dfrac{1}{2}\,\big|-36-16\big| = \dfrac{1}{2}(52) = 26$ ✓

(ii) Worked example — a second one

Find the area of the triangle with vertices $(5,-9)$, $(-4,3)$ and $(1,7)$.

Slide $(5,-9)$ to $(0,0)$ ($-5,\ +9$):

$(-4,3) \to (-9,12)$

$(1,7) \to (-4,16)$

$\text{Area} = \dfrac{1}{2}\,\big|\,(-9)(16) - (-4)(12)\,\big|$

$= \dfrac{1}{2}\,\big|-144 + 48\big| = \dfrac{1}{2}(96)$

$= 48$ square units

YOU TRY · 1

Find the area of the triangle with vertices $(0,0)$, $(4,1)$ and $(2,5)$.

A corner is already at the origin — straight into the formula.

$\dfrac{1}{2}\,\big|\,4(5) - 2(1)\,\big| = \dfrac{1}{2}(18)$

$9$ square units

YOU TRY · 2

Find the area of $\triangle ABC$ where $A(1,1)$, $B(4,2)$ and $C(2,5)$.

Slide $A$ to $(0,0)$ first — do the same to $B$ and $C$.

$-1,-1$: $B \to (3,1)$, $C \to (1,4)$.

$\dfrac{1}{2}\,\big|\,3(4) - 1(1)\,\big| = \dfrac{1}{2}(11)$

$5.5$ square units

Section 2 of 3

Collinear points

Collinear means the points sit on the same straight line.

Three collinear points can't make a triangle — so the "triangle" has area $0$.

Collinear test

1.slope of $AB$ = slope of $BC$, or

2.area of $\triangle ABC = 0$.

(i) Worked example

Show that $A(1,2)$, $B(3,6)$ and $C(5,10)$ are collinear.

Slide $A$ to $(0,0)$ ($-1,-2$): $B \to (2,4)$, $C \to (4,8)$.

$\text{Area} = \dfrac{1}{2}\,\big|\,2(8) - 4(4)\,\big| = \dfrac{1}{2}\,\big|16-16\big|$

Area $= 0 \Rightarrow$ the points are collinear.

YOU TRY · 3

Are $(0,0)$, $(2,3)$ and $(4,6)$ collinear?

Find the area. If it's $0$, they're collinear.

$\dfrac{1}{2}\,\big|\,2(6) - 4(3)\,\big| = \dfrac{1}{2}\,\big|12-12\big| = 0$

Yes — collinear.

Yes — area is $0$, so collinear.

Section 3 of 3

Finding an unknown

If you're given the area, you can work backwards to a missing coordinate.

(i) Worked example

$(1,-4)$, $(2,3)$ and $(5,t)$ form a triangle of area $12$ square units. Find the two values of $t$.

Slide $(1,-4)$ to $(0,0)$ ($-1,\ +4$):

$(2,3) \to (1,7)$

$(5,t) \to (4,\ t+4)$

So $(x_1,y_1) = (1,7)$, $(x_2,y_2) = (4,\ t+4)$.

$\dfrac{1}{2}\,\big|\,1(t+4) - 4(7)\,\big| = 12$

$\dfrac{1}{2}\,\big|\,t + 4 - 28\,\big| = 12$

$\big|\,t - 24\,\big| = 24$

Two cases:

$t - 24 = 24 \Rightarrow t = 48$

$t - 24 = -24 \Rightarrow t = 0$

$t = 48$ or $t = 0$

YOU TRY · 4

$(0,0)$, $(3,1)$ and $(t,4)$ form a triangle of area $5$ square units. Find the two values of $t$.

A corner is already at the origin. Set $\dfrac{1}{2}|\ldots| = 5$ and solve both cases.

$\dfrac{1}{2}\,\big|\,3(4) - t(1)\,\big| = 5 \Rightarrow \big|12 - t\big| = 10$

$12 - t = 10 \Rightarrow t = 2$; $12 - t = -10 \Rightarrow t = 22$.

$t = 2$ or $t = 22$

SUM

The lot in one box

Area toolkit

1.$\text{Area} = \dfrac{1}{2}\,\big|\,x_1 y_2 - x_2 y_1\,\big|$ — one corner at $(0,0)$.

2.Not at the origin? Slide the whole triangle so a corner lands on $(0,0)$.

3.Use the two non-origin corners as $(x_1,y_1)$ and $(x_2,y_2)$.

4.Collinear $\Leftrightarrow$ area $= 0$.

5.Given the area: set $\dfrac{1}{2}\,\big|\ldots\big| = A$ and solve the two cases.

End of lesson

Area of a Triangle — HL · Mathslive.ie