Question 1009850
THE EXPECTED WAY TO SOLVE IT:
You were taught the the distance from a point {{{P(x[0],y[0])}}}
to a line {{{ax+by+c=0}}} is measured along segment {{{PQ}}} ,
from {{{P}}} to the closest point on the line, point {{{Q}}} .
They also taught you a formula to find that distance:
{{{d(ax+by+c=0,P(x[0],y[0]))=abs(ax[0]+by[0]+c)/sqrt(a^2+b^2)}}} ,
and maybe even formulas to find the coordinates of point {{{Q}}} .
a) Lines parallel to {{{2x-3y=6}}}<-->{{{2x-3y-6=0}}}
Have the equation {{{2x-3y+c=0}}} .
Blindly applying that formula,
{{{d(2x-3y+c=0,P(-1,2))=abs(2(-1)-3(2)+c)/sqrt(2^2+(-3)^2)}}}
{{{d(2x-3y+c=0,P(-1,2))=abs(-2-6+c)/sqrt(4+9)}}}
{{{d(2x-3y+c=0,P(-1,2))=abs(c-8)/sqrt(13)}}} .
And since we know that distance must be {{{2sqrt(3)}}} ,
{{{abs(c-8)/sqrt(13)=2sqrt(3)}}}-->{{{abs(c-8)=2sqrt(3)*sqrt(13)}}}-->{{{abs(c-8)=2sqrt(39)}}}-->{{{c=8 +- 2sqrt(39)}}} .
So the equation of the two lines parallel to {{{2x-3y=6}}} ,
and passing at a distance {{{2sqrt(3)}}} units from {{{P(-1,2)}}} , is
{{{highlight(2x-3y+8 +- 3sqrt(39)=0)}}}
b) Line {{{2x-3y=6}}}<-->{{{2x-6=3y}}}<-->{{{y=(2/3)x-6/3}}}<-->{{{y=(2/3)x-2}}}
has a slope of {{{2/3}}} .
Lines perpendicular to {{{2x-3y=6}}}  have a slope of
{{{(-1)/(2/3)=(-1)*(3/2)=-3/2}}} .
Their equation will be
{{{y=-(3/2)x+b}}}<-->{{{2y=-3x+2b}}}<-->{{{3x+2y-2b=0}}} ,
which we could write as {{{3x+2y+c=0}}}.
As before
{{{d(3x+2y+c=0,P(-1,2))=abs(3(-1)+2(2)+c)/sqrt(3^2+2^2)}}}
{{{d(3x+2y+c=0,P(-1,2))=abs(-3+4+c)/sqrt(9+4)}}}
{{{d(3x+2y+c=0,P(-1,2))=abs(c+1)/sqrt(13)}}} ,
and since that distance must be {{{2sqrt(3)}}} ,
{{{abs(c+1)/sqrt(13)=2sqrt(3)}}}-->{{{abs(c+1)=2sqrt(3)*sqrt(13)}}}-->{{{abs(c+1)=2sqrt(39)}}}-->{{{c=-1 +- 2sqrt(39)}}} .
So the equation of the two lines perpendicular to {{{2x-3y=6}}} ,
and passing at a distance {{{2sqrt(3)}}} units from {{{P(-1,2)}}} , is
{{{highlight(3x+2y-1 +- 3sqrt(39)=0)}}} .


THE PICTURES AND EXPLANATION FOR THE SITUATION:
The line represented by {{{2x-3y=6}}} and
the circle that is the locus of all the points {{{2sqrt(3)}}} units from {{{P(-1,2)}}}
are shown below.
{{{drawing(300,300,-6,4,-4,6,grid(0),
circle(-1,2,0.1),locate(-1,2,P),
red(line(-6,-6,9,4)),
red(circle(-1,2,3.46)),
red(circle(-1,2,3.47))
)}}} The circle is centered at {{{p}}} and has a radius of {{{2sqrt(3)}}} units.The equation for the circle is {{{(x+1)^2+(y-2)^2=(2sqrt(3))^2}}}<-->{{{(x+1)^2+(y-2)^2=12}}} .
There is more than one line
parallel to {{{2x-3y=6}}} and passing at a distance {{{2sqrt(3)}}} units from {{{P(-1,2)}}} ,
and there is more than one line
perpendicular to 2x-3y=6 and passing at a distance {{{2sqrt(3)}}} units from {{{P(-1,2)}}} .
The way I read/interpret the problem, we want lines a) parallel, and b) perpendicular to {{{2x-3y=6}}} , and tangent to the circle {{{(x+1)^2+(y-2)^2=12}}}  .
Only {{{1}}} point of each of those lines is at a distance {{{2sqrt(3)}}} units from {{{P(-1,2)}}} .
That point is the intersection of the line with the circle {{{(x+1)^2+(y-2)^2=12}}}  .
The remaining point on each of the lines is at a distance greater than {{{2sqrt(3)}}} units from {{{P(-1,2)}}} ,
and therefore is outside the circle of radius.
 
There are two lines parallel to {{{2x-3y=6}}} ,
and tangent to the circle {{{(x+1)^2+(y-2)^2=12}}} .
They are tangent at points {{{Q}}} and {{{R}}} .
There are two lines perpendicular to {{{2x-3y=6}}} ,
and tangent to the circle {{{(x+1)^2+(y-2)^2=12}}}  .
They are tangent at points {{{S}}} and {{{T}}} .
{{{drawing(300,300,-6,4,-4,6,grid(0),
circle(-1,2,0.1),locate(-0.85,2.2,P),
red(line(-6,-6,9,4)),
red(circle(-1,2,3.46)),
red(circle(-1,2,3.47)),
circle(-3.882,0.078,0.1),
circle(1.882,3.922,0.1),
circle(-2.922,4.882,0.1),
circle(0.922,-0.882,0.1),
locate(1.08,-0.6,Q),locate(-2.922,5.4,R),
locate(-3.9,0.7,S),locate(2.032,4.122,T),
green(line(5,-7,-5,8)),
green(line(-7.882,6.078,2.118,-8.922)),
green(line(-2.118,9.922,7.882,-5.078)),
blue(line(-3.882,0.078,1.882,3.922)),
blue(line(-8.922,0.882,3.078,8.882)),
blue(line(-8.078,-6.882,6.922,3.118))
)}}} Each of those lines is perpendicular to the radius at the point of tangency.
So PQ and PR are perpendicular to the two lines parallel to {{{2x-3y=6}}} and must be perpendicular to {{{2x-3y=6}}} itself.
and PS and PT are perpendicular to the two lines perpendicular to {{{2x-3y=6}}} and must be parallel to {{{2x-3y=6}}} .