first off, let's keep in mind that perpendicular lines have negative reciprocal slopes, hmmm so what's the slope of y = 2x + 5?
well, notice, that equation is in slope-intercept form, thus [tex]\bf y=\stackrel{slope}{2}x+5[/tex].
[tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}}
{2\implies \stackrel{slope}{\cfrac{2}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{2}}}[/tex]
so, we're really looking for the equation of a line whose slope is -1/2 and runs through 1,4.
[tex]\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad \qquad \qquad
slope = m\implies -\cfrac{1}{2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-4=-\cfrac{1}{2}(x-1)[/tex]
