First of all, solve the equation #y-2x=5# for #y#

#y=color(green)2x+5#

The coefficient of #x# (#color(green)2#) is the slope, but it isn't the slope you are looking for, you are looking for the negative reciprocal for it since the line you are looking for is perpendicular to it.

#color(green)2# is the same as #color(green)2/1# so the slope you are looking for is, #m=color(blue)-1/color(green)2#

The equation until now is #y=-1/2x+b#

We need to find #b#, substitute the given point in this equation and find #b#, the point is #(color(red)1,color(blue)2)#, it corresponds to #(color(red)x,color(blue) y)#

#color(blue)y=-1/2color(red)x+b# #=># #color(blue)2=-1/2(color(red)1)+b#

#2=-1/2+b#

#b=2+1/2#

#b=4/2+1/2#

#b=5/2#

The equation is #y=-1/2x+5/2#