Question

If $A(2,4,-1), B(3,6,-1)$ and $C(4,5,1)$ are three consecutive vertices of a parallelogram, then its fourth vertex is

• A. (1,3,3)
• B. (1,3,-3)
• C. (3,3,-1)
• D. (3,3,1)

Answer 1

Answer: (D)

$A=(2,4,-1), B=(3,6,-1), C=(4,5,1)$
Let $D=(x, y, z)$
Since, diagonals of a parallelogram are bisect each other.
$\Rightarrow $ Mid-point of $A C=$ Mid-point of $B D$
$\left(\frac{2+4}{2}, \frac{4+5}{2}, \frac{-1+1}{2}\right)=\left(\frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2}\right)$
$\left(3, \frac{9}{2}, 0\right)=\left(\frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2}\right)$
$ \therefore \frac{3+x}{2}=3 ; \frac{6+y}{2}=\frac{9}{2} ; \frac{-1+z}{2}=0$
$3+x=6 ; 6+y=9 ;-1+z=0$
$x=3 ; y=3 ; z=1$
$\therefore $ Fourth vertex $D=(3,3,1)$