How do you write an equation that is perpendicular to y = -1/2x + 2/3, passes through (2,3)?

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Answer 1 · 2016-11-13T09:51:46

#y=2x-1#

Given:#" "y=-1/2x+2/3#

Compare to #y=mx+c# where #m# is the gradient (slope)

A straight line graph that is perpendicular to this would have the gradient of #-1/m#. So in this case #-1/m = +2/1 =2#

So we know that the line we are looking for is of form:

#y=2x+c#

This has 3 unknowns so we need to change it so that there is only 1 unknown and thus solvable.

We are told that it passes through the point #(x,y)->color(red)("(2,3)")# thus we have some known values. Lets substitute these:

#color(green)(y=2x+c" "->" "color(red)(3)=2(color(red)(2))+c)#

#3=4+c#

#c=-1#

So we now have #y=2x-1#