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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
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Verified Answer
The correct answer is:
$(3,3,1)$
$A=(2,4,-1), B=(3,6,-1), C=(4,5,1)$
Let $D=(x, y, z)$
Since, diagonals of a parallelogram are bisect each other.
$\Rightarrow$ Mid-point of $A C=$ Mid-point of $B D$
$$
\begin{aligned}
& \left(\frac{2+4}{2}, \frac{4+5}{2}, \frac{-1+1}{2}\right)=\left(\frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2}\right) \\
& \left(3, \frac{9}{2}, 0\right)=\left(\frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2}\right) \\
& \therefore \frac{3+x}{2}=3 ; \frac{6+y}{2}=\frac{9}{2} ; \frac{-1+z}{2}=0 \\
& 3+x=6 ; 6+y=9 ;-1+z=0 \\
& x=3 ; y=3 ; z=1
\end{aligned}
$$
$\therefore$ Fourth vertex $D=(3,3,1)$.
Hence, option (d) is correct.
Let $D=(x, y, z)$
Since, diagonals of a parallelogram are bisect each other.
$\Rightarrow$ Mid-point of $A C=$ Mid-point of $B D$
$$
\begin{aligned}
& \left(\frac{2+4}{2}, \frac{4+5}{2}, \frac{-1+1}{2}\right)=\left(\frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2}\right) \\
& \left(3, \frac{9}{2}, 0\right)=\left(\frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2}\right) \\
& \therefore \frac{3+x}{2}=3 ; \frac{6+y}{2}=\frac{9}{2} ; \frac{-1+z}{2}=0 \\
& 3+x=6 ; 6+y=9 ;-1+z=0 \\
& x=3 ; y=3 ; z=1
\end{aligned}
$$
$\therefore$ Fourth vertex $D=(3,3,1)$.
Hence, option (d) is correct.
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