Question

Find the coordinates of the intersection of the diagnosis of the parallelograms (-2,-1) (1,3) (6,3) and (3,-1)

Answer 1

Answer:

$M = (2,1)$

Step-by-step explanation:

Represent the diagonals as:

$A =(-2,-1)$

$B =(1,3)$

$C =(6,3)$

$D = (3,-1)$

Required

Determine the coordinate of the intersection

To do this, we simply calculate the midpoint of AC or BD.

For AC:

$(x_1,y_1) = (-2,-1)$

$(x_2,y_2) = (6,3)$

The midpoint is:

$M = \frac{1}{2}\{(x_1+x_2),(y_1+y_2)\}$

This gives:

$M = \frac{1}{2}\{(-2+6),(-1+3)\}$

$M = \frac{1}{2}\{(4),(2)\}$

$M = (\frac{1}{2} * 4,\frac{1}{2} * 2)$

$M = (2,1)$

For BD:

$(x_1,y_1) = (1,3)$

$(x_2,y_2) = (3,-1)$

The midpoint is:

$M = \frac{1}{2}\{(x_1+x_2),(y_1+y_2)\}$

This gives:

$M = \frac{1}{2}\{(1+3),(3-1)\}$

$M = \frac{1}{2}\{(4),(2)\}$

$M = (\frac{1}{2} * 4,\frac{1}{2} * 2)$

$M = (2,1)$

Notice the midpoints are the same:

$M = (2,1)$

Hence, the coordinates of the intersection is (2,1)