 # Find the coordinates of the point where the diagonals

Question:

Find the coordinates of the point where the diagonals of the parallelogram formed by joining the points (−2, −1), (1, 0), (4, 3) and(1, 2) meet.

Solution:

The co-ordinates of the midpoint $\left(x_{n}, y_{E}\right)$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by,

$\left(x_{w}, y_{m}\right)=\left(\left(\frac{x_{1}+x_{2}}{2}\right),\left(\frac{y_{1}+y_{2}}{2}\right)\right)$

In a parallelogram the diagonals bisect each other. That is the point of intersection of the diagonals is the midpoint of either of the diagonals.

Here, it is given that the vertices of a parallelogram are A(−2,−1), B(1,0) and C(4,3) and D(1,2).

We see that ‘AC’ and ‘BD’ are the diagonals of the parallelogram.

The midpoint of either one of these diagonals will give us the point of intersection of the diagonals.

Let this point be M(x, y).

Let us find the midpoint of the diagonal ‘AC’.

$\left(x_{w}, y_{m}\right)=\left(\left(\frac{-2+4}{2}\right),\left(\frac{-1+3}{2}\right)\right)$

$\left(x_{m}, y_{n}\right)=(1,1)$

Hence the co-ordinates of the point of intersection of the diagonals of the given parallelogram are $(1,1)$.