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

Concept: Area of Triangle by Coordinates — the area is half the absolute value of the determinant formed by the coordinates.

Step 1: Use the formula

Step 2: Substitute $(x_1,y_1)=(3,8)$, $(x_2,y_2)=(-4,2)$, $(x_3,y_3)=(5,1)$:

Step 3: Simplify inside:

Formula.

Substitute $(x_1,y_1)=(3,8),\ (x_2,y_2)=(-4,2),\ (x_3,y_3)=(5,1)$:

Check (vectors from $A(3,8)$). $\vec{AB}=(-7,-6),\ \vec{AC}=(2,-7)$:

Method: Area of a Triangle from Three Coordinate Points

This method finds the area of any triangle directly from its vertices' coordinates, without needing to find a base and height geometrically.

Steps

Step 1: Label the three vertices in order

Assign $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ to the three given points, in any consistent order (the formula works regardless of the order chosen, up to an overall sign that the absolute value removes).

Step 2: Apply the coordinate area formula

Notice the cyclic pattern: each $x_i$ multiplies the difference of the other two $y$-coordinates.

Step 3: Substitute and compute each bracket term

Work out $(y_2-y_3)$, $(y_3-y_1)$, $(y_1-y_2)$ first as plain numbers, then multiply each by its corresponding $x_i$.

Step 4: Sum the three terms, take the absolute value, then halve

Add the three products (watch negative signs carefully), take the absolute value of that sum (area is never negative), then multiply by $\tfrac12$.

Step 5: Sanity-check with the collinearity case

If the computed area comes out $0$, the three points are collinear, not a valid triangle — worth a quick mental check if the numbers look suspicious.

This coordinate formula is exact and works for any triangle orientation — never fall back to base-and-height geometry when coordinates are given directly.

Common Mistakes

Mistake 1: Forgetting the absolute value and reporting a negative area

Why it's wrong: the raw expression $x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$ can come out negative depending on the order the vertices are listed in — area itself can never be negative, so submitting a negative number as "the area" is a defect, not just a sign quirk. Correct approach: always take the absolute value of the bracketed sum before multiplying by $\tfrac12$.

Mistake 2: Dropping the factor of $\tfrac12$

Why it's wrong: the expression inside the absolute value bars is the area of a parallelogram (twice the triangle), not the triangle itself — forgetting to halve it doubles the final answer. Correct approach: always apply the $\tfrac12$ as the very last step, after taking the absolute value.