The three vertices of a parallelogram ABCD are A(3,−4),B(−1,−3) and C(−6,2). Find the coordinate of vertex D.
Let coordinates of vertex D be (x,y)
In parallelogram diagonals are bisect each other.∴ Mid-point of AC= Mid-point of BD
⇒ (3+(−6)2,−4+22)=(−1+x2,−3+y2)
⇒ (−32,−22)=(−1+x2,−3+y2)
⇒ (−32,−1)=(−1+x2,−3+y2)
Now,
⇒ −32=−1+x2
⇒ −6=−2+2x
⇒ −4=2x
∴ x=−2
⇒ −1=−3+y2
⇒ −2=−3+y
⇒ 1=y
∴ y=1
∴ Coordinates of vertex D is (−2,1)