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in 3D Coordinate Geometry by (4.0k points)
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Show that the points A(1, 3, 0), B(−5, 5, ), C(−9, −1, 2) and D(−3, −3,0) are the vertices of a parallelogram ABCD, but it is not a rectangle.

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AB =

\({\sqrt{{1+5)}^{2}+(3-5)^{2}+(0-2)^{2}}}\\=\sqrt{36+4+4}\\=\sqrt{44}\\=2\sqrt11 \)

BC =

\({\sqrt{{(-5+9)}^{2}+(5+1)^{2}+(2-2)^{2}}}\\=\sqrt{16+36+0}\\=\sqrt{52}\\=2\sqrt13 \)

CD =

\({\sqrt{{(-9+3)}^{2}+(-1+3)^{2}+(2-0)^{2}}}\\=\sqrt{36+4+4}\\=\sqrt{44}\\=2\sqrt11 \)

DA =

\({\sqrt{{(-3-1)}^{2}+(-3-3)^{2}+(0-0)^{2}}}\\=\sqrt{16+36+0}\\=\sqrt{42}\\=2\sqrt13 \)

Also, AC =

\({\sqrt{{(1+9)}^{2}+(3+1)^{2}+(0-2)^{2}}}\\=\sqrt{100+16+4}\\=\sqrt{120}\\=2\sqrt30 \)

Since AB =CD, BC = CA and AC ≠ BD

∴The opposite sides are equal and diagonals are unequal, so the given points are the vertices of a parallelogram not rectangle.

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