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Show that the points $A (1, 0), B (5, 3), C (2, 7)$ and $D (-2, 4)$ are the vertices of a parallelogram.
Given:
Given vertices are $A (1, 0), B (5, 3), C (2, 7)$ and $D (-2, 4)$.
To do:
We have to show that the points $A (1, 0), B (5, 3), C (2, 7)$ and $D (-2, 4)$ are the vertices of a parallelogram.
Solution:
We know that,
Diagonals of a parallelogram bisect each other.
Let the diagonals $AC$ and $BD$ bisect each other at $O(x,y)$.
This implies, using mid-point formula, we get,
\( O(x,y)=\left(\frac{1+2}{2}, \frac{0+7}{2}\right) \)
\( =\left(\frac{3}{2}, \frac{7}{2}\right) \)
\( O \) is the mid-point of \( B D, \) then,
\( O(x,y)=\left(\frac{5+(-2)}{2}, \frac{3+4}{2}\right) \)
\( =\left(\frac{5-2}{2}, \frac{3+4}{2}\right) \)
\( =\left(\frac{3}{2}, \frac{7}{2}\right) \)
Coordinates of mid-point of $AC =$ Coordinates of mid-point of $BD$
Therefore, the points $A (1, 0), B (5, 3), C (2, 7)$ and $D (-2, 4)$ are the vertices of a parallelogram.
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