Area of a Triangle by Determinant

From earlier classes, the triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ has area $\frac{1}{2}\left[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right]$. That expression is exactly a determinant expansion, so the formula packs into

$$\Delta = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$$

— one row per vertex, a column of $1$s on the right.

Three remarks carry all the marks in this section:

1. Area is positive — always take the absolute value of the determinant. A negative determinant just means the vertices were listed clockwise.
2. When the area is given, work with both signs: set $\frac{1}{2}(\text{determinant}) = \pm\,\text{area}$. Solving with only the $+$ sign silently throws away half the answers — the single most common error here.
3. Collinear points give zero area. Three points lie on one line exactly when the determinant vanishes — a clean collinearity test with no slopes to compare.

Equation of a line through two points

Remark 3 doubles as a line-finding tool. A point $P(x, y)$ lies on the line through $A$ and $B$ exactly when the triangle $ABP$ is flat:

$$\begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} = 0$$

Expanding gives the equation of the line directly — no slope formula, no point-slope juggling, and it never breaks on vertical lines.

Solved Examples

Example 1 — A straight area computation

Find the area of the triangle whose vertices are $(3, 8)$, $(-4, 2)$ and $(5, 1)$.

Step 1 — set up the determinant: $$\Delta = \frac{1}{2}\begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & 1 & 1 \end{vmatrix}$$

Step 2 — expand along $R_1$: $$\Delta = \frac{1}{2}\left[3(2 - 1) - 8(-4 - 5) + 1(-4 - 10)\right] = \frac{1}{2}(3 + 72 - 14)$$

Answer: $\Delta = \dfrac{61}{2}$ square units.

Example 2 — Line through two points, then a vertex hunt

Find the equation of the line joining $A(1, 3)$ and $B(0, 0)$ using determinants, and find $k$ if $D(k, 0)$ makes the area of triangle $ABD$ equal to $3$ square units.

Step 1 — line as a flat triangle: for any point $P(x, y)$ on $AB$, the triangle $ABP$ has zero area: $$\frac{1}{2}\begin{vmatrix} 0 & 0 & 1 \\ 1 & 3 & 1 \\ x & y & 1 \end{vmatrix} = 0$$

Step 2 — expand: $\frac{1}{2}(y - 3x) = 0$, so the line is $y = 3x$.

Step 3 — impose the given area (both signs!): $$\frac{1}{2}\begin{vmatrix} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{vmatrix} = \pm 3 \quad\Rightarrow\quad \frac{-3k}{2} = \pm 3$$

Answer: the line is $y = 3x$ and $k = \mp 2$ — both $k = 2$ and $k = -2$ work, one on each side of the line.

Example 3 — Given area, find the unknown vertex

Find the values of $k$ if the area of the triangle with vertices $(k, 0)$, $(4, 0)$, $(0, 2)$ is $4$ square units.

Step 1 — set up with $\pm$: $$\frac{1}{2}\begin{vmatrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = \pm 4$$

Step 2 — expand along $C_2$ (two zeros): the only nonzero entry in $C_2$ is the $2$ at position $(3,2)$, sign $(-1)^{3+2} = -1$: $$\frac{1}{2}\left[-2\begin{vmatrix} k & 1 \\ 4 & 1 \end{vmatrix}\right] = \frac{1}{2}\left[-2(k - 4)\right] = 4 - k$$

Step 3 — solve both cases: $4 - k = 4$ gives $k = 0$; $4 - k = -4$ gives $k = 8$.

Answer: $k = 0$ or $k = 8$ — two triangles of area $4$, one with the moving vertex on each side.

Example 4 — The collinearity test

Show that the points $A(a, b+c)$, $B(b, c+a)$, $C(c, a+b)$ are collinear.

Step 1 — area determinant: $$\Delta = \frac{1}{2}\begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix}$$

Step 2 — expand and watch it cancel: $$2\Delta = a\left[(c+a) - (a+b)\right] - (b+c)(b - c) + 1\left[b(a+b) - c(c+a)\right]$$ $$= a(c - b) - (b^2 - c^2) + ab + b^2 - c^2 - ac = ac - ab - b^2 + c^2 + ab + b^2 - c^2 - ac = 0$$

Answer: the determinant vanishes, so the area is $0$ and the three points are collinear. (Notice each row sums to $a + b + c$ in its first two entries plus the constant $1$ — a structural reason for the collapse that Section 8's properties will make instant.)