Area of a Triangle from Its Vertices

For the area of a triangle (or any polygon) from coordinates, the shoelace formula is faster and safer than splitting into shapes. Lay the vertices out, apply the pattern, and take the modulus so orientation cannot spoil the sign.

The triangle $ABC$ has vertices $A(1, 1)$, $B(5, 2)$ and $C(3, 6)$. Find its area. [4]

The working

Use the triangle area formula: $\text{Area} = \frac{1}{2}\left| x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B) \right|$

Substitute $A(1,1)$, $B(5,2)$, $C(3,6)$ (M1 for the setup): $= \frac{1}{2}\left| 1(2 - 6) + 5(6 - 1) + 3(1 - 2) \right|$

Evaluate each bracket (M1): $= \frac{1}{2}\left| 1(-4) + 5(5) + 3(-1) \right| = \frac{1}{2}\left| -4 + 25 - 3 \right| \quad \text{(A1)}$

Finish: $= \frac{1}{2}\left| 18 \right| = 9 \text{ square units} \quad \text{(A1)}$

Where the marks are won and lost

• Keep the vertices in a consistent cycle ($A \to B \to C$) so the differences pair correctly. Jumbling the order corrupts the sum.
• The modulus guards the answer: if the vertices happen to be listed clockwise, the inside comes out negative, and the modulus fixes it. Never report a negative area.
• Each term is $x_i \times (\text{difference of the other two } y\text{s})$. Mismatching an $x$ with the wrong $y$-difference is the usual slip.

Common mistakes

• Dropping the modulus and reporting a negative area.
• Pairing $x_A$ with $(y_A - y_B)$ instead of $(y_B - y_C)$.
• Forgetting the overall factor of $\frac12$.

Full method: Area of Rectilinear Figures notes. Topic home: Straight-Line Graphs pillar.