we know the line runs through -2,3 and 2,5.
it also runs through 2, 5 and 6,k.
since all points are on the line, they're colinear, and therefore the line runs also through -2,3 and 6,k.
keeping in mind that line maintains a constant slope, therefore, the slope for -2,3 and 2,5, has to be the same slope as for 2,5 and 6,k.
what is the slope of -2,3 and 2,5 anyway?
[tex]\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~{{ -2}} &,&{{ 3}}~)
% (c,d)
&&(~{{ 2}} &,&{{ 5}}~)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}\implies
\cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{5-3}{2-(-2)}\implies \cfrac{5-3}{2+2}\implies \cfrac{2}{4}\implies \cfrac{1}{2}[/tex]
and since we know the slope of 2,5 and 6,k is the same, then,
[tex]\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~{{ 2}} &,&{{ 5}}~)
% (c,d)
&&(~{{ 6}} &,&{{ k}}~)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}\implies
\cfrac{\stackrel{rise}{{{ y_2}}-{{ y_1}}}}{\stackrel{run}{{{ x_2}}-{{ x_1}}}}\implies \cfrac{k-5}{6-2}\implies \cfrac{k-5}{4}=\stackrel{slope}{\cfrac{1}{2}}
\\\\\\
2k-10=4\implies 2k=14\implies k=\cfrac{14}{2}\implies k=7[/tex]