Question

A Square has the vertices at (-2,6) (6,1) (1,-7) (-7,-2). At what point do the diagonals of the square intercect

Answer 1

Answer: At point ($-\frac{1}{2},-\frac{1}{2}$)

Step-by-step explanation: Diagonal is a line uniting two opposite points. In a square, the diagonals intersect in a 90° and bisect each other, i.e., divides each diagonal into two segments of the same length.

In other words, the diagonals of a square meet at their midpoint, which is found as the following:

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

The opposite vertices of the given square are (-2,6) and (1,-7).

So, the intersection is

(x,y) = $(\frac{-2+1}{2} ,\frac{6-7}{2})$

(x,y) = $(-\frac{1}{2},-\frac{1}{2} )$

The diagonals of square with vertices (-2,6)(6,1)(1,-7)(-7,-2) intersect at point $(-\frac{1}{2},-\frac{1}{2} )$.