Area of a Triangle / Collinearity

MATArea & Collinearity

The area of a triangle with given vertices can be written as a determinant, which makes collinearity simply the condition that this determinant equals zero. The formula carries an absolute value because area is non-negative, and dropping it is the most common slip.

ISC examines this as short questions: compute an area, or find a parameter $k$ so that three points are collinear or enclose a stated area.

Area as a determinant

\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|

(x_1,y_1),(x_2,y_2),(x_3,y_3)

are the vertices of the triangle; the outer bars denote absolute value.

Collinearity condition

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

Three points are collinear iff the area is zero, i.e. the determinant vanishes (no

\tfrac12

or modulus needed here).

Equation of a line through two points

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

Setting the area of

(x,y),(x_1,y_1),(x_2,y_2)

to zero gives the line joining the two fixed points.

Place the three coordinates as rows with a final column of $1$ s, expand the $3\times3$ determinant, multiply by $\tfrac12$ , and take the modulus for area.
Expanding along the last column is fastest since each entry is $1$ .
For collinearity set the bare determinant to $0$ ; the factor $\tfrac12$ and the absolute value are irrelevant to the equation.
When a problem fixes the area (e.g. exactly $4$ square units), solve $\tfrac12\,|\Delta(k)| = 4$ , which gives $\Delta(k) = \pm 8$ and hence two possible values of $k$ .
The sign of the unmodulused determinant indicates the orientation (anticlockwise positive, clockwise negative) of the vertex ordering.
Units of area are the coordinate units squared.
Simplify the determinant expression in $k$ fully before equating, to avoid losing a root.

Where the marks go

Omitting the absolute value and reporting a negative area.
Keeping the $\tfrac12$ or modulus when writing the collinearity equation $\Delta = 0$ , which over-complicates the algebra.
When an area is prescribed, solving only $\Delta = +8$ and missing the $\Delta = -8$ case, so one valid value of $k$ is lost.
Entering coordinates in the wrong columns (swapping $x$ and $y$ ) and getting a sign or value error in the determinant.

How the board asks it

Numericalarea as a determinant
Find the area of the triangle whose vertices are $A(2,7)$ , $B(1,1)$ and $C(10,8)$ , using a determinant.
Numericalcollinearity condition
Find the value of $k$ for which the points $(k,2-2k)$ , $(-k+1,2k)$ and $(-4-k,6-2k)$ are collinear.
Derive / provebare determinant equals zero
Using determinants, show that the points $(a,b+c)$ , $(b,c+a)$ and $(c,a+b)$ are collinear.
Numericalprescribed area gives two values
If the area of the triangle with vertices $(-2,0)$ , $(0,4)$ and $(0,k)$ is $4$ square units, find the value(s) of $k$ .
Numericalequation of a line through two points
Using a determinant, find the equation of the line joining the points $(1,2)$ and $(3,4)$ .

Practise this topic

PreviousSolving a System of Linear Equations (Matrix Method) & Consistency

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.