Area of a Triangle Using Determinants

The formula

For a triangle with vertices (x1,y1)(x_1,y_1), (x2,y2)(x_2,y_2) and (x3,y3)(x_3,y_3), the area is half the absolute value of a determinant:

Area=12∣∣x1y11x2y21x3y31∣∣\text{Area} = \dfrac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|

The determinant can be negative, depending on the order the vertices are listed in (it is positive for anticlockwise order), but an area never is, hence the absolute value.

Worked example

Find the area of the triangle with vertices (2,7)(2,7), (1,1)(1,1) and (10,8)(10,8).

∣2711111081∣=2(1−8)−7(1−10)+1(8−10)=−14+63−2=47\begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix} = 2(1-8) - 7(1-10) + 1(8-10) = -14 + 63 - 2 = 47

So the area is 472\dfrac{47}{2} square units.

Worked example: testing for collinearity

Are (1,2)(1,2), (3,6)(3,6) and (5,10)(5,10) collinear?

∣1213615101∣=1(6−10)−2(3−5)+1(30−30)=−4+4+0=0\begin{vmatrix} 1 & 2 & 1 \\ 3 & 6 & 1 \\ 5 & 10 & 1 \end{vmatrix} = 1(6-10) - 2(3-5) + 1(30-30) = -4 + 4 + 0 = 0

A zero area means the three points lie on one line, here y=2xy = 2x.

Worked example: an unknown vertex

The triangle with vertices (k,0)(k,0), (4,0)(4,0) and (0,2)(0,2) has area 4. Find kk.

12∣k(0−2)+4(2−0)+0(0−0)∣=4  ⇒  ∣8−2k∣=8\dfrac{1}{2}\left|k(0-2) + 4(2-0) + 0(0-0)\right| = 4 \;\Rightarrow\; |8-2k| = 8

So 8−2k=88 - 2k = 8 or 8−2k=−88 - 2k = -8, giving k=0k = 0 or k=8k = 8. Both work: the base along the xx-axis is 4 units either way, and the height is 2.

Why the order of the vertices changes the sign

Listing the vertices of the first example in the order (2,7)(2,7), (10,8)(10,8), (1,1)(1,1) swaps two rows of the determinant, which flips its sign: it becomes −47-47. Anticlockwise order gives a positive value and clockwise a negative one, which is why the formula takes the absolute value before halving.

Worked example: a quadrilateral

Find the area of the quadrilateral with vertices (0,0)(0,0), (4,0)(4,0), (5,3)(5,3) and (1,4)(1,4), taken in that order.

Solution: Split it along the diagonal from (0,0)(0,0) to (5,3)(5,3) into two triangles and apply the formula to each:

12∣∣001401531∣∣=6,12∣∣001531141∣∣=172\dfrac{1}{2}\left|\begin{vmatrix} 0 & 0 & 1 \\ 4 & 0 & 1 \\ 5 & 3 & 1 \end{vmatrix}\right| = 6, \qquad \dfrac{1}{2}\left|\begin{vmatrix} 0 & 0 & 1 \\ 5 & 3 & 1 \\ 1 & 4 & 1 \end{vmatrix}\right| = \dfrac{17}{2}

The total is 6+172=2926 + \dfrac{17}{2} = \dfrac{29}{2} square units. The split only works along a diagonal that lies inside the shape, which it does here because the quadrilateral is convex.

Common mistakes

  • Dropping the absolute value. A negative determinant still gives a positive area.
  • Forgetting the 12\tfrac{1}{2}. The determinant alone is twice the area.
  • Keeping only one solution when the area is given: ∣D∣=c|D| = c has two cases, D=cD = c and D=−cD = -c.
  • Using this in 3D. For points with three coordinates, use the cross product instead.

Where this comes up in exams

JEE Main 2026 lists the area of triangles using determinants in its Matrices and Determinants unit, and the 2025-26 CBSE Class XII syllabus lists applications of determinants in finding the area of a triangle.

Checked against JEE (Main) 2026 syllabus (NTA) and CBSE Mathematics curriculum, Classes XI-XII (2025-26). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

Continue learning

For triangles in three dimensions the cross product does the same job.

From past GATE papers