As the given coordinates are:
A(1,4), B(2,2) and C(-2,1)
And the given scalar facto is: K=6
So The new coordinates are:
[tex]\begin{gathered} A^{\prime}=(1\times6,\text{ 4}\times6)=(6,\text{ 24)} \\ B^{\prime}=(2\times6,\text{ 2}\times6)=(12,\text{ 12)} \\ C^{\prime}=(-2\times6,\text{ 1}\times6)=(-12.,\text{ 6)} \end{gathered}[/tex]
And the diagram for the new coordinates is:
