Question 132228
Lines are perpendicular if and only if their slopes are negative reciprocals of each other.  If {{{m[1]}}} is the slope of one of the lines and {{{m[2]}}} is the slope of the other line, then they are perpendicular if {{{m[1]=-1/m[2]}}}.


Start with the equation of the given line and put it into slope-intercept form ({{{y=mx+b}}}) by solving for y:


{{{-7x-8y=77}}}


{{{-8y=7x+77}}}


{{{y=-7x/8-77/8}}}


Therefore the slope of the given line is {{{-7/8}}} because that is the coefficient on x when the equation is in slope-intercept form.


Since you want to derive the equation for a line perpendicular to the given line, you need the negative reciprocal of the slope of the given line, so:


{{{-1/(-7/8)=8/7}}}


Now that you have the slope determined and you are given a point, you can use the point-slope form to derive the equation of the desired perpendicular.


Point-slope form: {{{y-y[1]=m(x-x[1])}}} where ({{{x[1]}}},{{{y[1]}}}) are the co-ordinates of the given point and m is the slope.


{{{y-7=(8/7)(x-(-3))}}}, is an equation for the desired line.  However, you probably want to put it into standard form just like the given equation at the start.


{{{y-7=(8/7)(x+3)}}}


Multiply by 7
{{{7y-49=8x+24}}}


{{{-8x+7y=24+49}}}


{{{-8x+7y=73}}}