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