Show that the points A (1, 0), B (5, 3),


Show that the points A (1, 0), B (5, 3), C (2, 7) and D (−2, 4) are the vertices of a parallelogram.


Let A (1, 0); B (5, 3); C (2, 7) and D (−2, 4) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram.

We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram.

Now to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as,

$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

So the mid-point of the diagonal AC is,

$\mathrm{Q}(x, y)=\left(\frac{1+2}{2}, \frac{0+7}{2}\right)$

$=\left(\frac{3}{2}, \frac{7}{2}\right)$

Similarly mid-point of diagonal BD is,

$R(x, y)=\left(\frac{5-2}{2}, \frac{3+4}{2}\right)$

$=\left(\frac{3}{2}, \frac{7}{2}\right)$

Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other.

Hence ABCD is a parallelogram.

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