Question 1013618
Reference equation given, {{{y=(2/3)x-4}}}.  Slope is {{{2/3}}}.
A line in the plane parallel also has slope  {{{2/3}}}; and a line in the plane perpendicular to the reference line has slope negative reciprocal, which is  {{{-3/2}}}.


The parallel line through the given point (-2,-5) is  {{{y-(-5)=(2/3)(x-(-2))}}} which is written in point-slope form.


The perpendicular line through the given point is  {{{y-(-5)=(-3/2)(x-(-2))}}}, again being in point-slope form.


Starting those in point-slope form was easier than arranging for slope-intercept form.  Use your algebra skill to put each of them into standard form.