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 $AC=$ Mid-point of $BD$
$\left(\frac{2+4}{2},\frac{4+5}{2},\frac{-1+1}{2}\right)=\left(\frac{3+x}{2},\frac{6+y}{2},\frac{-1+z}{2}\right)$
$\left(3,\frac{9}{2},0\right)=\left(\frac{3+x}{2},\frac{6+y}{2},\frac{-1+z}{2}\right)$
$\therefore \frac{3+x}{2}=3;\frac{6+y}{2}=\frac{9}{2};\frac{-1+z}{2}=0$
$3+x=6;6+y=9;-1+z=0$
$x=3;y=3;z=1$
$\therefore$ Fourth vertex $D=(3,3,1)$