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.

Given: 

The given points are $(-2, -1), (1, 0), (4,3)$ and $(1, 2)$.

To do: 

We have to 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:

Let the vertices of the given parallelogram be $A (-2, -1), B (1, 0), C (4, 3)$ and $D (1, 2)$ in which $AC$ and $BD$.

We know that,

Diagonals of a parallelogram bisect each other.

Let $O(x,y)$ be the point of intersection of $AC$ and $BD$.

This implies, \( \mathrm{O} \) is the mid-point of \( \mathrm{AC} \).

Using the mid-point formula,

$(x,y)=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Therefore,

The Coordinates of the point \( \mathrm{O} \) are
\( O(x,y)=\left(\frac{-2+4}{2}, \frac{-1+3}{2}\right) \)

\( =\left(\frac{2}{2}, \frac{2}{2}\right) \)

\( =(1,1) \)

The coordinates of the point where the diagonals meet is $(1,1)$.

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Updated on: 10-Oct-2022

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