Split it into two separate instructions: the magnitude $|k|$ controls size (image distances from the centre are $|k|$ times the object's), and the negative sign controls position — the image sits on the opposite side of the centre from the object, which makes it look rotated $180°$. A positive $k$ keeps the image upright on the same side; a negative $k$ flips it through the centre.

For each vertex: draw a ray from the vertex through the centre and beyond, measure the distance from the centre to the vertex, multiply by $|k|$, then plot the image that new distance from the centre, on the far side of the ray.

Enlarge a triangle by $k=-2$ from centre $C$, where one vertex $A$ is $5$ cm from $C$: new distance $=5\times2=10$ cm, plotted from $C$ along the ray through $A$, but on the opposite side of $C$.

Draw a straight line through each pair of corresponding vertices ($A$ to $A'$, $B$ to $B'$); their single intersection point is the centre $C$ (a third pair is a good check). For the scale factor, compare corresponding lengths, and remember the sign comes from the inversion:

Every point's "journey" from the centre scales by $k$:

Enlarge $P(2,1)$ by $k=-2$ from $C(1,0)$: journey $=(2-1,\,1-0)=(1,1)$; scaled $=(-2,-2)$; image $=(1+(-2),\,0+(-2))=(-1,-2)$.

Finding an unknown centre and $k$: if $A(4,5)\to A'(1,2)$ and $B(2,3)\to B'(5,6)$, set up $1-x=k(4-x)$ and $5-x=k(2-x)$ from the $x$-components (and similarly for $y$), then solve simultaneously — subtracting the two $x$-equations eliminates $x$ and gives $k$ directly, which you then substitute back to find the centre.

Area always uses $k^2$, and squaring a negative number gives a positive result — so the area scale factor is unaffected by the sign: $(-3)^2=9$, same as $(+3)^2$.

Volume uses $k^3$, and a negative number cubed stays negative — a volume of $-80\text{ cm}^3$ doesn't mean a literal negative volume, it's a bookkeeping signal that the solid has been inverted through the centre; the actual volume magnitude is still $80\text{ cm}^3$, scaled by $|k|^3$.

Shape A is enlarged with scale factor $k=-1.5$, centre the origin, to give shape B. (a) Describe fully the single transformation mapping B back to A. (b) The area of B is $63\text{ cm}^2$ — find the area of A.

1. (a) The inverse of any enlargement has the reciprocal scale factor and the same centre: enlargement, scale factor $\dfrac{1}{-1.5}=-\dfrac{2}{3}$, centre the origin.
2. (b) Area scale factor $=(-1.5)^2=2.25$. Since $\text{Area}_B = \text{Area}_A \times 2.25$, $\text{Area}_A = \dfrac{63}{2.25}=28\text{ cm}^2$.

The sign of $k$ never affects an area calculation — only its magnitude, squared.