Question 200525
Take note that the slope is {{{m=2}}}, which can be written as {{{m=2/1}}}. Flip this fraction and change the sign to get {{{m=-1/2}}}. Now it turns out that any equation of the form {{{y=-(1/2)x+b}}} (where "b" is a constant) is perpendicular to {{{y=2x+3}}}.


So the equations {{{y=-(1/2)x+2}}}, {{{y=-(1/2)x+1/3}}}, {{{y=-(1/2)x+55}}}, {{{y=-(1/2)x+11}}}, etc.. are all perpendicular to {{{ y= 2x +3}}}