Question

Show that the
points (4,7,8) (2,3,4),(-1,-2,1),(1,2,5) are vertices
of Parallelogram​

Answer 1

AB= CD

DC= AD

BC not equal to DB

do ABCD is a parallelogram

Answer 2

Given:

Vertices = (4,7,8) (2,3,4),(-1,-2,1),(1,2,5)

To prove that:

The given vertices are the vertices of a parallelogram.

Solution:

$Direction ratios of a line passing through two points P\left(x_{1}, y_{1}, z_{1}\right),\& Q\left(x_{2}\right.&\left.y_{2}, z_{2}\right) \\\\&=\left(x_{2}-x_{1}\right),\left(y_{2}-y_{1}\right),\left(z_{2}-z_{1}\right).$

$Let A \equiv(4,7,8), B \equiv(2,3,4), C \equiv(-1,-2,1) and D \equiv(1,2,5) .$

Direction ratios of parallel lines are equal.

$Direction ratios of A B=\left(\frac{2}{6}, \frac{4}{6}, \frac{4}{6}\right) Direction ratios of C D=\left(\frac{-2}{6}, \frac{-4}{6}, \frac{-4}{6}\right)=\left(\frac{2}{6}, \frac{4}{6}, \frac{4}{6}\right) So, A B is parallel to CD.$

$Direction ratios of A D=\left(\frac{3}{\sqrt{43}}, \frac{5}{\sqrt{43}}, \frac{3}{\sqrt{43}}\right) Direction ratios of B C=\left(\frac{-3}{\sqrt{43}}, \frac{-5}{\sqrt{43}}, \frac{-3}{\sqrt{43}}\right)=\left(\frac{3}{\sqrt{43}}, \frac{5}{\sqrt{43}}, \frac{3}{\sqrt{43}}\right) So, A D is parallel to B C .$

The opposite sides are parallel and congruent in a parallelogram. So, ABCD is a parallelogram.

Therefore, the  points (4,7,8) (2,3,4),(-1,-2,1),(1,2,5) are the vertices  of a parallelogram​.