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?
![\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}]()
and since we know the slope of 2,5 and 6,k is the same, then,
![\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]()