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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Aicharya · Accepted Answer

$A=(2,4,-1), B=(3,6,-1), C=(4,5,1)$ 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}ight)=\left(\frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2}ight) \ & \left(3, \frac{9}{2}, 0ight)=\left(\frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2}ight) \ & 	herefore \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} $ $	herefore$ Fourth vertex $D=(3,3,1)$. Hence, option (d) is correct

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

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