Question 797588
EDIT: USE OF GRAPH GAVE THE CORRECT ANSWER.  See Below.


.... at least you should put A, B, and C on graph paper, and decide where to put D.  Then you can use simple coordinate geometry of lines and the distance formula, maybe, to find the coordinates needed for D.  Since you want a parallelogram, you know that there are two pair of parallel lines, you know two points define any line...



The question seems slightly open-ended.  Arbitrarily picking BC as one of the sides, and AB as the other side, the two sides meeting at point B, the slope for the side BC is {{{(-1-3)/(4-1)=-4/3}}}.


The parallelogram needs another side, AD, which has the same slope, -4/3, as line BC.  Note too that line AD obviously contains point A.   We have the coordinates for point A, being (-2,4).  We can use the POINT-SLOPE form for a linear equation to get line AD.


Line AD is {{{y-v=m(x-u)}}}, where m is slope, (u,v) is a point known on the line.
{{{y-4=(-4/3)(x-(-2))}}}
{{{y-4=-(4/3)x-8/3}}}
{{{y=-(4/3)x+4-8/3}}}-------(MISTAKE STEP HERE.  SHOULD HAVE BEEN {{{highlight(y=-(4/3)x+12/3-8/3)}}}, and then {{{highlight(y=-(4/3)x+4/3)}}})
{{{y=-(4/3)x-4/3}}}, line AD.


You also want to know line DC, because this will intersect line AD, at point D.  Line DC will need to be parallel to line AB, so we must find slope for AB, and use known point C in POINT-SLOPE form for a line to find line DC.


Slope AB is slope DC, {{{(3-4)/(1-(-2))=-1/3}}}, this slope.
{{{y-v=m(x-u)}}} which is generalized point-slope form; see above.
{{{y-(-1)=-(1/3)(x-4)}}}
{{{y=-(1/3)x+4/3 -1}}}
{{{y=-(1/3)x+1/3}}}, line DC.


FINALLY,  at what point D, do lines AD and DC intersect?
Equating their formulas for y,
{{{-(4/3)x-4/3=-(1/3)x+1/3}}}
{{{-4x-4=-x+1}}}
{{{-4=4x-x+1}}}
{{{-5=3x}}}
{{{highlight(x=-5/3)}}}


Use either equations to find y from the now found x:
DC, {{{y=-(1/3)(-5/3)+1/3}}}
{{{y=5/6+1/3}}}
{{{y=5/6+2/6=7/6}}}
{{{highlight(y=7/6)}}}


May possibly be a mistake somewhere there, but again, use a gridded sheet of graph paper to check and make a good visual estimate, at least.  My calculations here seem to indicate {{{x=-1&2/3}}}, {{{y=1&1/6}}}


----------------------------
A mistake MUST have been made in the above work.  I just tried the points on graph paper, and it seems, point D should be at (1,0).  Try my advice generally, but be careful, and suspicious of the values I picked and calculated.  My graph seems to show, slope for AB is -1/3, and slope for BC is -4/3; so those much are good.


BRIEF SUMMARY OF SOLUTION METHOD SUCCESSFUL:   Tried again using symbolism and algebra.  Better.  Slope AB  is -1/3.
Slope of BC is -4/3.  


Point Slope form was used next.  General point (u,v), form is y-v=m(x-u).


Line AD parallel to BC.  Line AD uses slope -4/3 and point A(-2,4).  Resulting equation in standard form is 4x+3y=4.


Line DC parallel to AB.  Line DC uses slope -1/3 and point C(4,-1).  Resulting equation in standard form is x+3y=1.


Elimination method on the system of standard form equations gives x=1 and y=0.  
PERFECT.   Point is D(1,0).