Question 951372
P (-1,-4) Q ( 1,1) R (4,2)

{{{drawing( 600, 600, -10, 10, -10, 10,

circle(-1,-4,.12),circle(1,1,.12),circle(4,2,.12),
locate(-1,-4,P(-1,-4)),locate(1,1,Q(1,1)),locate(4,2,R(4,2)),
 graph( 600, 600, -10, 10, -10, 10,0)) }}}

if the figure {{{PQRS}}} is a parallelogram, the point S should be located at intersection point of the line parallel to PQ and the line parallel to SR 

to find the line parallel to PQ , first find a line PQ:




{{{-4=m(-1) + b}}}
{{{-4=-m + b}}}...eq.1
{{{1=m + b}}}...eq.2
-----------------------add
{{{-4+1=-m+m+b + b}}}
{{{-3=2b}}}
{{{-3/2=b}}}
{{{-1.5=b}}}
find {{{m}}}
{{{1=m + b}}}...eq.2
{{{1=m -1.5}}}
{{{1+1.5=m }}}
{{{2.5=m }}}

the equation of this line is:
{{{y=2.5x-1.5}}}

the line {{{SR}}} has same slope, so:

{{{y=2.5x + b}}}...since this line passes through  R (4,2), we know that {{{2=2.5*4 + b}}}

{{{2=10 + b}}}

{{{2-10=b}}}

{{{b=-8}}}

the equation of this line is:
{{{y=2.5x-8}}}
{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-1,-4,.12),circle(1,1,.12),circle(4,2,.12),
locate(-1,-4,P(-1,-4)),locate(1,1,Q(1,1)),locate(4,2,R(4,2)),
 graph( 600, 600, -10, 10, -10, 10,2.5x -1.5,2.5x-8)) }}}

same way we will find the line SR parallel  to QR
 the line QR: if Q ( 1,1) R (4,2)

{{{2=m(4) + b}}}
{{{2=4m + b}}}...eq.1
{{{1=m + b}}}...eq.2
-----------------------subtract

{{{2-1=4m-m + b-b}}}

{{{1=3m}}}
{{{m=1/3}}}

{{{1=(1/3) + b}}}...eq.2

{{{1-1/3=b}}}

{{{b=2/3}}}

the equation of this line is:

{{{y=(1/3)x+2/3}}}

parallel line is:{{{y=(1/3)x+b}}} and passes through P (-1,-4)

{{{-4=(1/3)*(-1)+b}}}

{{{-4=-1/3+b}}}

{{{-4+1/3=b}}}

{{{b=-11/3}}}

{{{y=(1/3)x-11/3}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-1,-4,.12),circle(1,1,.12),circle(4,2,.12),
locate(-1,-4,P(-1,-4)),locate(1,1,Q(1,1)),locate(4,2,R(4,2)),
 graph( 600, 600, -10, 10, -10, 10,2.5x -1.5,2.5x-8,(1/3)x+2/3,(1/3)x-11/3)) }}}


intersection of {{{y=(1/3)x+2/3}}} and  {{{(1/3)x-11/3}}} is the point S

from the graph, I see the coordinates of the point S are (2,-3)

check if it is true

{{{2.5x-8=(1/3)x-11/3}}}

{{{7.5x-24=x-11}}}

{{{7.5x-x=24-11}}}

{{{6.5x=13}}}

{{{x=13/6.5}}}

{{{x=2}}}

and {{{y=2.5*2-8=5-8=-3}}}

so,  the point {{{S}}} is at  ({{{2}}},{{{-3}}})


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-1,-4,.12),circle(1,1,.12),circle(4,2,.12),circle(2,-3,.12),
locate(-1,-4,P(-1,-4)),locate(1,1,Q(1,1)),locate(4,2,R(4,2)),locate(2,-3,S(2,-3)),
 graph( 600, 600, -10, 10, -10, 10,2.5x -1.5,2.5x-8,(1/3)x+2/3,(1/3)x-11/3)) }}}