Question

Points A(-2, 4), B(1, 3), C(4, -1) and D form a parallelogram. What are the coordinates of D? (5, 5) (0, 0) (1, -2) (1, 0) (3, 4)

Answer 1

If its a ABCD parallelogram ,
then, let , D(x,y)
so DA=$\sqrt{(-2+x)^{2} +(4+y)^{2} }$ 
BC=$\sqrt{(1+4)^{2}+(3-1)^{2} }$
    =$\sqrt{25+4}$
    =$\sqrt{29}$
According to question , 
$\sqrt{(-2+x)^{2} +(4+y)^{2} }$=$\sqrt{29}$    
[Because, DA||BC]
or, (x-2)²+(y+4)²=29.....(1)
 Again,
CD=$\sqrt{(x+4)^{2} +(y-1)^{2} }$  
and AB=$\sqrt{(-2+1)^{2}+(4+3)^{2} }$
           =$\sqrt{1+49}$
           =$\sqrt{50}$
According to question,
(x+4)²+(y-1)²=50.......(2)
[Because,CD||AB]
From (1),
x²-4x+4+y²+8y+16=29
x²+y²-4x+8y=29.....(3)
from (2),
x²+8x+16+y²-2y+1=50
x²+y²+8x-2y=33.....(4)
From(3)-(4),
-12x+10y=-4
or,-6x+5y=-2
or,5y=6x-2
or,y=$\frac{6x-2}{5}$.....(5)
Again (3)+(4),
4x+6y=62
or,2x+3y=31
or,2x+$\frac{3(6x-2)}{5}$=31
or,$\frac{10x+18x-6}{5}$=31
or,28x-6=155
or,28x=161
or,x=$\frac{161}{28}$ =5.75
By putting x=$\frac{161}{28}$ in (5) ,
y=$\frac{ \frac{6X161}{28} -2 }{5}$ 
or,y=$\frac{ \frac{966}{28} -2 }{5}$
or,y=$\frac{ \frac{966-56}{28} }{5}$
or,y=$\frac{940}{140}$
or,y=6.5 
So D(x,y)=(5.75,6.5)