# Verify that $(-1,2,1),\:(1,-2,5),\:(4,-7,8)\:and\:(2,-3,4)$ are vertices of a parallelogram.

## 1 Answer

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• distance between the points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is given by$\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Let $A(-1,2,1),\:B(1,-2,5),\:C(4,-7,8)\:and\:D(2,-3,4)$ be the given four points.
We know that the distance between the points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is given by
$\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Distance $\overline {AB}=\sqrt {(1+1)^2+(-2-2)^2+(5-1)^2}=\sqrt{4+16+16}=\sqrt {36}=6$
Distance $\overline {BC}=\sqrt {(4-1)^2+(-7+2)^2+(8-5)^2}=\sqrt{9+25+9}=\sqrt {43}$
Distance $\overline {CD}=\sqrt {(2-4)^2+(-3+7)^2+(4-8)^2}=\sqrt{4+16+16}=\sqrt {36}=6$
Distance $\overline {AD}=\sqrt {(2+1)^2+(-3-2)^2+(4-1)^2}=\sqrt{9+25+9}=\sqrt {43}$
$\Rightarrow\:\overline {AB}=\overline {CD}$ and $\overline {BC}=\overline {AD}$
$\therefore$ The four points are form a parallelogram.
answered Apr 17, 2014
edited Apr 17, 2014

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