Question 1153113
The coordinates of three vertices of a parallelogram are 

A({{{-3}}}, {{{3}}}), B({{{2}}}, {{{3}}}),=> these two points lie on a horizontal line {{{y=3}}}

 and C({{{4}}}, {{{-1}}}), D({{{x}}}, {{{y}}})=> since parallelogram,these two points must lie on a horizontal line {{{y=-1}}}

so far we know D({{{x}}}, {{{-1}}})

since parallelogram, {{{AB=CD}}}; so, the distance between them is same

find the distance between points {{{A}}} and {{{B}}}

since both points lie on horizontal line, count the number of units between them

{{{d=5}}}

or, you can do it this way:

{{{d=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}

{{{d=sqrt((2-(-3))^2+(3-3)^2)}}}

{{{d=sqrt((2+3)^2+(0)^2)}}}

{{{d=sqrt(5^2)}}}

{{{d=5}}}

now, you know the distance, y coordinate, and point C: 

 {{{d=5}}},{{{y=-1}}},

C({{{4}}}, {{{-1}}})

D({{{x}}}, {{{-1}}})

since point A in quadrant II, point D in quadrant III because {{{AD||BC}}}

go from C({{{4}}}, {{{-1}}}) 5 units from {{{x=4}}} to the left and you will et to {{{-1}}}

=> the fourth vertex is at D({{{-1}}}, {{{-1}}})



{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(-3,3,.12), locate(-3,3.5,A),
circle(2,3,.12), locate(2,3.5,B),
circle(4,-1,.12), locate(4,-1.5,C),
circle(-1,-1,.12), locate(-1.2,-1.5,D), 
green(line(-1,-1,-3,3)),green(line(-3,3,2,3)),
green(line(4,-1,2,3)),
graph( 600, 600, -10, 10, -10, 10, 3,3, -1)) }}}