Area of triangle using coordinate formula: Worked examples

This page is a workout. The parent note built the formula and proved it. Here we throw every kind of triangle at it — friendly ones, ugly ones, broken ones — so that no exam question can surprise you.

Everything below uses the single tool from the parent:

Before any numbers, a reminder of what the symbols mean, from zero:

A vertex is just a corner point, written as a pair (across, up) — how far right and how far up from the origin.
A negative coordinate means "the other way": $x = - 3$ is 3 steps left, $y = - 3$ is 3 steps down.
"Square units" is the unit of area — how many $1 \times 1$ tiles fit inside.

The scenario matrix

Every triangle an exam can hand you falls into one of these boxes. The right column names the worked example that kills that box.

#	Case class	What's tricky about it	Example
1	All-positive, generic triangle	Baseline — just plug in	Ex 1
2	Mixed / negative coordinates	Sign bookkeeping in the products	Ex 2
3	Right triangle on the axes	Should match $\frac{1}{2} bh$	Ex 3
4	Degenerate — collinear points	Formula gives 0	Ex 4
5	Clockwise vs counter-clockwise order	Inside goes negative; $∣ \cdot ∣$ rescues it	Ex 5
6	Solve backwards for an unknown coordinate	Set Area $=$ target, solve for the letter	Ex 6
7	Real-world word problem (a field / plot)	Translate words $\to$ points, carry units	Ex 7
8	Exam twist — quadrilateral by splitting	Two triangles, add the areas	Ex 8

Case 1 — the plain triangle

Worked example Example 1 · all-positive vertices

(matrix cell 1) Given: $A (1, 2), B (4, 6), C (5, 1)$ . Find the area.

Forecast: eyeball it — the points span about 4 across and 5 up, so a triangle should be under half of $4 \times 5 = 20$ . Guess "around 9–10". Write your guess down.

Step 1. Label the slots: $x_{1} = 1, y_{1} = 2; x_{2} = 4, y_{2} = 6; x_{3} = 5, y_{3} = 1$ . Why this step? The formula only works if each number sits in the correct box; a mislabel poisons everything after.

Step 2. Build each $x_{i} \times (other two y ’s)$ : $x_{1} (y_{2} - y_{3}) = 1 (6 - 1) = 5, x_{2} (y_{3} - y_{1}) = 4 (1 - 2) = - 4, x_{3} (y_{1} - y_{2}) = 5 (2 - 6) = - 20$ Why this step? This is the cyclic pattern in action — do all three separately so a sign error is easy to spot.

Step 3. Sum and halve with bars: $Area = \frac{1}{2} ∣5 - 4 - 20∣ = \frac{1}{2} ∣ - 19∣ = \frac{19}{2} = 9.5$ Why this step? The bars convert the "backwards-order" minus into a clean positive area.

Verify: $9.5$ sits right inside our forecast of $9$ – $10$ . ✓

Case 2 — negative coordinates

Case 3 — right triangle on the axes

Figure — Area of triangle using coordinate formula

Case 4 — collinear (the triangle that isn't)

Case 5 — orientation: clockwise vs counter-clockwise

Worked example Example 5 · same triangle, reversed order

(matrix cell 5) Given: the triangle $P (0, 0), Q (4, 0), R (0, 3)$ . Compute the signed value (no bars yet) once listed counter-clockwise $P \to Q \to R$ , and once clockwise $P \to R \to Q$ .

Forecast: it's a right triangle, legs 4 and 3, so $∣ Area ∣ = \frac{1}{2} (4) (3) = 6$ . Guess that the two orderings give $+ 6$ and $- 6$ .

Step 1 — counter-clockwise $P, Q, R$ : signed $= \frac{1}{2} [0 (0 - 3) + 4 (3 - 0) + 0 (0 - 0)] = \frac{1}{2} (12) = + 6$ . Why this step? Dropping the bars keeps the sign so we can see orientation. Walking anticlockwise gives $+$ .

Step 2 — clockwise $P, R, Q$ : signed $= \frac{1}{2} [0 (3 - 0) + 0 (0 - 0) + 4 (0 - 3)] = \frac{1}{2} (- 12) = - 6$ . Why this step? Same three points, order flipped — the sign flips too. The green (CCW) and coral (CW) arrows in the figure show the two walking directions.

Step 3. Take absolute value in both: $∣ + 6 ∣ = ∣ - 6 ∣ = 6$ . Why this step? Area is a size, not a direction — this is why the parent formula wears the bars.

Verify: both give $6$ , equal to $\frac{1}{2} \cdot 4 \cdot 3 = 6$ . ✓ The sign is orientation information, discarded for area (kept in the Vector Cross Product and Determinants).

Case 6 — solve backwards for a missing coordinate

Worked example Example 6 · unknown vertex from a target area

(matrix cell 6) Given: $A (1, 1), B (5, 1), C (k, 4)$ , and the area is required to be $6$ square units. Find all values of $k$ .

Forecast: base $A B$ is horizontal, length 4, and $C$ is 3 units above that line — so height is 3 and area is $\frac{1}{2} (4) (3) = 6$ for any $k$ ? Guess: $k$ is free... let's test that suspicion.

Step 1. Set up the formula with $k$ inside: $Area = \frac{1}{2} 1 (1 - 4) + 5 (4 - 1) + k (1 - 1)$ Why this step? Keep $k$ symbolic so we can solve, not just evaluate.

Step 2. Simplify — the $k$ term dies: $= \frac{1}{2} ∣ - 3 + 15 + 0 ∣ = \frac{1}{2} (12) = 6$ Why this step? Because $B$ and $A$ share the same $y = 1$ , the coefficient of $k$ is $(y_{1} - y_{2}) = 0$ . So $k$ drops out.

Step 3. Conclusion: the area is $6$ for every real $k$ . Why this step? $C$ slides along the horizontal line $y = 4$ ; its height above base $A B$ stays $3$ , so area never changes.

Verify: try $k = 0$ : $\frac{1}{2} ∣1 (- 3) + 5 (3) + 0∣ = \frac{1}{2} ∣12∣ = 6$ . Try $k = 100$ : $\frac{1}{2} ∣ - 3 + 15 + 100 (0) ∣ = 6$ . ✓ Both hit the target — the forecast was right.

Case 7 — real-world word problem

Worked example Example 7 · a triangular plot of land

(matrix cell 7) Given: A park's triangular flower bed has survey corners at $A (2, 3), B (8, 3), C (6, 9)$ , measured in metres. Fertiliser covers $4 m^{2}$ per bag. How many whole bags are needed?

Forecast: the plot looks maybe $6$ m wide and $6$ m tall, so area is a good fraction of $18$ . Guess "around 18 m², so about 5 bags".

Step 1. Area in square metres: $Area = \frac{1}{2} ∣2 (3 - 9) + 8 (9 - 3) + 6 (3 - 3) ∣$ Why this step? Coordinates are in metres, so the formula returns square metres directly — units come for free.

Step 2. $= \frac{1}{2} ∣2 (- 6) + 8 (6) + 6 (0) ∣ = \frac{1}{2} ∣ - 12 + 48 + 0∣ = \frac{1}{2} (36) = 18 m^{2}$ Why this step? Straight arithmetic once the substitution is clean.

Step 3. Bags $= \frac{18}{4} = 4.5$ , and you can't buy half a bag $\Rightarrow$ round up to $5$ bags. Why this step? Real quantities are discrete; under-buying leaves bare soil, so we ceil.

Verify: base $A B = 8 - 2 = 6$ m, height $= 9 - 3 = 6$ m, $\frac{1}{2} (6) (6) = 18 m^{2}$ . ✓ Matches, units are m², answer $= 5$ bags.

Case 8 — exam twist: area of a quadrilateral

Worked example Example 8 · split a quadrilateral into two triangles

(matrix cell 8) Given: quadrilateral $P (0, 0), Q (6, 0), R (7, 4), S (2, 5)$ taken in order. Find its area.

Forecast: it's a slanted four-sided patch, spread about 7 across and 5 up. Guess "somewhere near 30".

Step 1. Draw the diagonal $P R$ , splitting the shape into $△ P QR$ and $△ P R S$ (see the dashed butter diagonal in the figure). Why this step? The coordinate formula only knows triangles; any polygon is just triangles glued along a diagonal.

Step 2 — $△ P QR$ with $P (0, 0), Q (6, 0), R (7, 4)$ : $\frac{1}{2} ∣0 (0 - 4) + 6 (4 - 0) + 7 (0 - 0) ∣ = \frac{1}{2} ∣24∣ = 12$ Why this step? One triangle, one plug-in; nothing new.

Step 3 — $△ P R S$ with $P (0, 0), R (7, 4), S (2, 5)$ : $\frac{1}{2} ∣0 (4 - 5) + 7 (5 - 0) + 2 (0 - 4) ∣ = \frac{1}{2} ∣35 - 8∣ = \frac{27}{2} = 13.5$ Why this step? Reuse the same diagonal $P R$ so the two pieces exactly tile the quadrilateral with no overlap.

Step 4. Add: $12 + 13.5 = 25.5$ square units. Why this step? Areas of non-overlapping pieces simply add.

Verify: the Shoelace Theorem on all four points in order gives $\frac{1}{2} ∣ (0 \cdot 0 - 6 \cdot 0) + (6 \cdot 4 - 7 \cdot 0) + (7 \cdot 5 - 2 \cdot 4) + (2 \cdot 0 - 0 \cdot 5) ∣ = \frac{1}{2} ∣0 + 24 + 27 + 0∣ = \frac{1}{2} (51) = 25.5. ✓$ Same $25.5$ , close to the forecast of "near 30". ✓

Recall Quick self-test on the cases

A triangle gives area $0$ from the formula — what does that mean? ::: The three points are collinear; no triangle exists (matrix cell 4). Listing vertices clockwise instead of counter-clockwise changes what? ::: Only the sign of the un-barred value; the area (after $∣ \cdot ∣$ ) is unchanged (cell 5). Why did the unknown $k$ vanish in Example 6? ::: Its base was horizontal ( $y_{1} = y_{2}$ ), so $C$ 's height above it — and thus the area — is fixed no matter where $C$ slides. How do you get the area of a 4-sided polygon with this triangle tool? ::: Split it along a diagonal into two triangles and add their areas (cell 8).