# Area of a Triangle in Coordinates

> Area of a triangle in coordinates for JEE Mathematics: determinant formula, collinearity test, vector cross product route and the shoelace extension.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/area-triangle-coordinate
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Area of a Triangle in Coordinates", PrepElephant, https://prepelephant.com/topics/jee/mathematics/area-triangle-coordinate

## Direct answer

Half the absolute value of one determinant: for vertices (x1, y1), (x2, y2), (x3, y3), the area is ½|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)| — and the same expression collapsing to zero is the collinearity test, one formula answering two question types. With one vertex at the origin it reduces to ½|x1y2 − x2y1|, the two-dimensional shadow of the vector route ½|AB × AC|. Polygons extend the idea through the shoelace arrangement: list vertices in boundary order, multiply along diagonals, take the difference, halve. The outer modulus means vertex order never matters for a triangle, and ½ × base × height wins whenever a horizontal or vertical base sits in the data.

## What you must remember

- **The determinant:** ½|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)| — cyclic structure, so any rotation of the vertices gives the same number.
- **Collinearity:** the bracketed expression equal to zero means the three points lie on one line; area and collinearity are the same question twice.
- **Origin shortcut:** with one vertex at (0, 0), area = ½|x1y2 − x2y1| — the cross product of the two position vectors.
- **Vector route:** ½|AB × AC| for the triangle in space or plane; it doubles as the cross-product chapter's triangle area.
- **Shoelace for polygons:** with vertices listed in order, area = ½|Σ(xi·yi+1 − xi+1·yi)| — the order of listing matters even though the modulus hides orientation.
- **Median split:** each median divides the triangle into two equal areas, and the centroid trisects every median.
- **Base-height survival:** ½ × base × height wins whenever a side is horizontal or vertical — read the height straight off the coordinates.

## Two routes to one area

Compute the area of the triangle with vertices (2, 3), (−1, 0) and (2, −4). The determinant route: ½|2(0 − (−4)) + (−1)(−4 − 3) + 2(3 − 0)| = ½|8 + 7 + 6| = 21/2 = 10.5 square units. The vector route should agree: from (2, 3), the side vectors are (−3, −3) toward (−1, 0) and (0, −7) toward (2, −4); their scalar cross product is (−3)(−7) − (−3)(0) = 21, and half its absolute value is again 21/2. Two independent routes, one answer — the cheapest error-check in coordinate geometry, and worth performing on every high-stakes area computation. Notice also the structural read: two vertices share x-coordinate 2, so the vertical base has length 7 and the horizontal distance of (−1, 0) from that line is 3; base-height gives ½ × 7 × 3 = 21/2 instantly. Three methods, ranked by the data's shape — the skill is choosing before computing, not memorising one and forcing it.

## Where marks leak

JEE Main asks the determinant area and the collinearity condition, and the two recurring losses are the missing ½ (options carry double the area as the top distractor) and sign slips when expanding the bracketed expression — writing x1(y2 − y3) as x1y2 − x1y3 requires the discipline of keeping parentheses until the end. A subtler Main pattern hides collinearity inside "find k so the points are collinear", which is the determinant set to zero and solved — the same formula wearing a parameter. JEE Advanced builds the triangle from lines instead of points: three pairwise intersections of lines give the vertices, then the determinant runs — and the entire question tests whether the intersections were solved correctly first. Another Advanced face puts one vertex on a curve at a parameter and asks when the area is minimal, welding this topic to differentiation. The shoelace extension demands boundary order: vertices listed in a scrambled order produce a self-crossing polygon whose shoelace value is not the visual area, so reorder before expanding, never after.

## Frequently asked questions

### What is the area formula for a triangle from its vertices?

½|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)| — one cyclic expansion with the modulus outside.

### How does the formula test collinearity?

Three points are collinear exactly when the bracketed expression vanishes — zero area and one line are the same statement.

### What is the quickest formula when one vertex is the origin?

½|x1y2 − x2y1| for the other two vertices — the scalar cross product halved.

### How do you find the area of a polygon with many vertices?

Use the shoelace sum ½|Σ(xi·yi+1 − xi+1·yi)| with the vertices listed in boundary order, closing the loop back to the first.

### When is base times height the better method?

Whenever a side is horizontal or vertical, the height is read directly from the other coordinate — as with the 21/2 example's vertical base of length 7.