Show that the quadrilateral whose vertices are (2, -1) (3, 4), (-2, 3) and (-3, -2) is a rhombus.
Let A(2, -1), B(3, 4), C(-2, 3) and D(-3, -2)

![](https://www.sarthaks.com/?qa=blob&qa_blobid=11453103522098884174)

Since AB=BC=CD=AD

Therefore, ABCD is a rhombus
solution: ` x = {:((-4/5,-1),(2/5,1)):}` answer: `(x-1)/(3)=(y+2)/(-4) =(z-3)/(5) , vec( r) =(hat(i) -2hat(j) +3hat(k)) +lambda (3hat(i) -4hat(j) +5hat(k))` let abc be the trianglewhere ab=(a-1)2 cmbc=2√a cmca=(a+1) cmab2=(a-1)2=a2+1-2abc2=(2√a)2=4a2ca2=(a+1)2=a2+1+2ahence ab2+bc2=ac2so Δabc is a right angles triangle at b a - ii, b - iv, c - i, d - iii quadrille, quadruple, quadruplet, quadrupled, quadruples, kibbles. vertical and horizontal multiply 2 in both lhs and rhs,integrating on both the sides, the distance to be covered to go one round along perimeter of this quadrilateral is12 cm + 3 cm + 5.6 cm + 19 cm solution: `abs(veca)=abs(vecb)=abs(vecc)=lamda`   

`veca*vecb=vecb*vecc*vecc*veca=0`   

`abs(veca+vecb+vecc)^2=(veca+vecb+vecc)*(veca+vecb+vecc)`   

`=abs(veca)^2+0+0+abs(vecb)^2+abs(vecc)^2`   

`=lambda^2+lambda^2+lambda^2`   

`=3lambda^2`   

`abs(veca+vecb+vecc)=sqrt3lambda`   

`costheta\_1=(veca*(veca+vecb+vecc))/(abs(veca)abs(veca+vecb+vecc))=lambda^2/(lambdasqrt3lambda)=1/sqrt3=theta\_1=cos^(-1)(1/sqrt3)`   

similarly,   

`theta\_1=theta\_2=theta\_3=cos^(-1)(1/sqrt3)` if the figure below represents a parabola, identity the physical&nbsp;quantities