Area of a triangle using determinants

Give three points on a coordinate plane and ask for the area of the triangle they form, and there is a shoelace-style formula from coordinate geometry that will get you there — but it is clumsy, sign-sensitive, and easy to misremember. Determinants package the same computation into one clean, symmetric object: write the three vertices as rows of a 3×33\times3 array, evaluate its determinant, take half its absolute value, and the area falls out. The same array does double duty. Set it equal to zero instead of evaluating it for area, and it becomes an exact test for whether three points are collinear — the triangle they "form" has degenerated to a straight line. This single idea, from the CBSE Class 12 chapter on Determinants, is one of the two named applications of determinants (alongside solving linear equations via the inverse matrix), and it shows up constantly wherever coordinate geometry and linear algebra meet.

The core insight

The area of a triangle with vertices (x1,y1)(x_1,y_1), (x2,y2)(x_2,y_2), (x3,y3)(x_3,y_3) equals half the absolute value of the 3×33\times3 determinant formed by writing each vertex as a row (xi,yi,1)(x_i, y_i, 1), and this same determinant equals zero exactly when the three points are collinear.

The core idea

Setting up the array. Each vertex contributes one row to a 3×33\times3 determinant, and every row carries the same pattern: the xx-coordinate, then the yy-coordinate, then a fixed 11.

D=∣x1y11x2y21x3y31∣D = \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}

where: (x1,y1)(x_1,y_1), (x2,y2)(x_2,y_2), (x3,y3)(x_3,y_3) are the three vertices in the order you choose to list them, and DD is the resulting signed number — the raw determinant, before it is turned into an area.

Why absolute value is unavoidable. An area can never be negative, yet DD can come out negative depending purely on the order the vertices are listed in — swapping any two rows flips the sign of a determinant. So the working formula always wraps DD in absolute value bars before halving:

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

where: the vertical bars around the full array denote "take the absolute value of the number the determinant evaluates to," applied only after expansion, not to each entry individually.

The collinearity test. Because area is zero exactly when a "triangle" has flattened into a line, three points (x1,y1)(x_1,y_1), (x2,y2)(x_2,y_2), (x3,y3)(x_3,y_3) are collinear precisely when

∣x1y11x2y21x3y31∣=0\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0

with no absolute value or halving needed here, since we only care whether the value is exactly zero, not what it equals otherwise.

Figure
Two rows of a determinant array turned into signed triangle strips, with three collinear points shown flattening the triangle to a line of zero area
Same determinant, two readings: area, or a test for a straight line genuine triangle, area not zero D not equal 0, Area = half of D collinear points, area zero D = 0, triangle flattens to a line
Derivation AHalf the base times the height, seen as a determinant

Show that the determinant formula Area=12∣∣x1y11x2y21x3y31∣∣\text{Area} = \tfrac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right| reduces to the familiar half-base-times-height rule, so that the determinant is not a mysterious new object but the same geometric area written compactly.

Step 1. Take one side of the triangle, say the segment joining (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2), as the base, and let the perpendicular distance from (x3,y3)(x_3,y_3) to the line through the first two points be the height hh. The base length is b=(x2−x1)2+(y2−y1)2b = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.

Step 2. The equation of the line through (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2) can be written as (y2−y1)(x−x1)−(x2−x1)(y−y1)=0(y_2-y_1)(x-x_1) - (x_2-x_1)(y-y_1) = 0, so the perpendicular distance from (x3,y3)(x_3,y_3) to this line is h=∣(y2−y1)(x3−x1)−(x2−x1)(y3−y1)∣(x2−x1)2+(y2−y1)2h = \dfrac{\left|(y_2-y_1)(x_3-x_1) - (x_2-x_1)(y_3-y_1)\right|}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}.

Step 3. Multiply base by height and halve: Area=12 b h=12∣(y2−y1)(x3−x1)−(x2−x1)(y3−y1)∣\text{Area} = \tfrac12 \, b \, h = \tfrac12 \left|(y_2-y_1)(x_3-x_1) - (x_2-x_1)(y_3-y_1)\right|, since the square-root factor in bb cancels the one in hh.

Step 4. Expand the bracket: (y2−y1)(x3−x1)−(x2−x1)(y3−y1)=x1y2−x1y3+x2y3−x2y1+x3y1−x3y2(y_2-y_1)(x_3-x_1) - (x_2-x_1)(y_3-y_1) = x_1y_2 - x_1y_3 + x_2y_3 - x_2y_1 + x_3y_1 - x_3y_2.

Step 5. Now expand the determinant DD along its third column (all entries 11) using cofactors: D=1⋅(x2y3−x3y2)−1⋅(x1y3−x3y1)+1⋅(x1y2−x2y1)D = 1\cdot(x_2y_3-x_3y_2) - 1\cdot(x_1y_3-x_3y_1) + 1\cdot(x_1y_2-x_2y_1), which is exactly the same six-term expression obtained in Step 4, only reordered. Hence Area=12∣D∣\text{Area} = \tfrac12|D|, confirming the determinant formula is just half-base-times-height in disguise, packaged so no separate base or height ever needs to be computed by hand.

Worked example

Find the area of the triangle with vertices (1,0)(1,0), (6,0)(6,0), and (4,3)(4,3), then use the same determinant idea to check a second set of points for collinearity.

Step 1. Set up the determinant with (x1,y1)=(1,0)(x_1,y_1)=(1,0), (x2,y2)=(6,0)(x_2,y_2)=(6,0), (x3,y3)=(4,3)(x_3,y_3)=(4,3):

∣101601431∣\begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix}

Step 2. Expand along the second column, since two of its three entries are already zero. Only the third-row entry, 33, survives, with cofactor sign (−1)3+2=−1(-1)^{3+2}=-1:

D=−3∣1161∣=−3[(1)(1)−(1)(6)]=−3(−5)=15D = -3\begin{vmatrix}1&1\\6&1\end{vmatrix} = -3\big[(1)(1)-(1)(6)\big] = -3(-5) = 15

Step 3. Apply the area formula: Area=12∣15∣=7.5\text{Area} = \tfrac12|15| = 7.5 square units.

Step 4. Now check whether A(3,−2)A(3,-2), B(−1,4)B(-1,4), C(5,−8)C(5,-8) are collinear by forming the same determinant and expanding along the first row:

∣3−21−1415−81∣=3[(4)(1)−(1)(−8)]+2[(−1)(1)−(1)(5)]+1[(−1)(−8)−(4)(5)]\begin{vmatrix} 3 & -2 & 1 \\ -1 & 4 & 1 \\ 5 & -8 & 1 \end{vmatrix} = 3\big[(4)(1)-(1)(-8)\big] + 2\big[(-1)(1)-(1)(5)\big] + 1\big[(-1)(-8)-(4)(5)\big]

Step 5. Evaluate each bracket: 3(12)+2(−6)+1(−12)=36−12−12=123(12) + 2(-6) + 1(-12) = 36 - 12 - 12 = 12. Since the determinant is 12≠012 \neq 0, the points are not collinear, and if they were joined into a triangle its area would be 12(12)=6\tfrac12(12) = 6 square units.

Competitive stretch

If instead the determinant for A(3,−2)A(3,-2), B(−1,4)B(-1,4), C(5,k)C(5,k) is given to be zero, the same setup gives a linear equation in kk: expanding along the first row as before, with y3=ky_3=k now appearing in two of the three minors, yields 3(4−k)+2(−1−5)+1(−k−20)=03(4-k) + 2(-1-5) + 1(-k-20) = 0, i.e. 12−3k−12−k−20=012 - 3k - 12 - k - 20 = 0, so −4k=20-4k = 20 and k=−5k = -5 — the determinant-equals-zero trick doubles as a way to solve for an unknown coordinate that forces collinearity.

Common traps

  • Forgetting the absolute value before halving — reporting a "negative area" or skipping straight from DD to the final answer without the 12∣⋅∣\tfrac12|\cdot| step.
  • Mixing up which column holds xx, which holds yy, and which holds the constant 11 — every row must follow the same (xi,yi,1)(x_i, y_i, 1) pattern, in that order.
  • Treating a small non-zero determinant as "basically collinear" — only an exact value of 00 proves collinearity; anything else is a genuine, if thin, triangle.
  • Dropping the factor of 12\tfrac12 and reporting the raw determinant value as the area itself.

Ask the tutor: if you have three coordinate points and want the area or a collinearity check worked through step by step, ask the AI tutor to set up the determinant with you.

Does the order in which I list the three vertices in the determinant matter

It changes the sign of DD but never its magnitude, so it never changes the final area. Listing (x1,y1),(x2,y2),(x3,y3)(x_1,y_1),(x_2,y_2),(x_3,y_3) versus swapping any two rows flips DD from positive to negative or back, which is exactly why the formula insists on the absolute value before halving. Pick any order that is convenient for expansion — the answer is the same either way.

Can I use this same determinant for points in 3D space

No, not directly — this exact 3×33\times3 layout with a column of ones is built for points given as plane coordinates (x,y)(x,y), and it computes ordinary two-dimensional area. Points in three dimensions need a different tool, typically the cross product of two side vectors, which is beyond this chapter's scope; here, stick to coordinate-plane triangles.

If the determinant comes out as a fraction, does that mean I made a mistake

Not necessarily — vertices with fractional or decimal coordinates, or an odd combination of integer coordinates, can legitimately produce a fractional area, just as example 1 above gave 7.57.5 square units from whole-number vertices. Double-check the row setup and cofactor signs before assuming an error; a fractional final answer is often correct.

Before this

Where this leads