Question 120206
First step is to find the required slope value.  Perpendicular lines have slopes that are negative reciprocals of each other:  {{{m[1]=-(1/m[2])}}}


The equation of the perpendicular is {{{2x+7y=16}}}.  Put this into slope-intercept form, {{{y=mx+b}}}, by solving for y.


{{{2x+7y=16}}}
{{{7y=-2x+16}}}
{{{y=((-2)/7)x +16}}}, which tells us that the slope of the perpendicular to the desired line is {{{-(2/7)}}}.


Using {{{m[1]=-(1/m[2])}}}, we now know that the slope of the desired line is {{{7/2}}}.


Now we have a point, (5,-3) and a slope, {{{7/2}}}, so use the point slope form of the line:


{{{(y-y[1])=m(x-x[1])}}} where m is the slope and ({{{x[1]}}},{{{y[1]}}}) are the coordinates of the given point.


{{{y-(-3)=(7/2)(x-5))}}}


Now simplify to standard form:


Multiply by 2


{{{2y+6=7(x-5)}}}


Distribute:


{{{2y+6=7x-35}}}


Add -7x to both sides


{{{-7x+2y+6=-35}}}


Add -6 to both sides:


{{{-7x+2y=-41}}} is the desired equation, though it would be a little tidier to multiply through by -1:


{{{7x-2y=41}}}


{{{graph(600,600,-20,20,-20,20,grid(2),-(2x/7)+16,(7/2)x-(41/2))}}}
Hope this helps,
John