Given the coordinates
)
and
)
you know that the distance

and you know that

must be the diagonal of the square (it can't be a side as

is horizontal but the sides of the square have gradients of

and

).
Now, for any square of side length

with diagonal length

, Pythagoras' Theorem gives us that

, and since the area of a square is

, we can conclude that:
and since we know here that

, it follows that:
If you were not told that the two normals and tangents enclosed a square, you could deduce it geometrically as follows:
- The angle between the tangent at
and the normal at
is a right angle from the definition of a normal at a point on a curve as the line perpendicular to the tangent at that point.
- It follows that the angle between the tangent at
and the normal at
is also a right angle
- We know that the tangents at
and
have gradients of
and
, respectively, and since
, these two tangents are perpendicular
- So, our quadrilateral is either a square or a rectangle (as all four of its angles are 90 degrees)
- The midpoint of
is at
, and it must also be the midpoint of the other diagonal,
, whether the quadrilateral is a square or a rectangle
- If the shape is a square, the diagonals must be perpendicular, and since
is horizontal, that would make
vertical, so let's try for
and
to lie on 
- Now, the tangent at
has a gradient of 1 and so any change in
will be accompanied by the same change in
- that is 
- Starting from
, we need
to move to
, making
, and putting
at  = \Big(7\cfrac{1}{2} - \cfrac{1}{2}, 3\cfrac{1}{4} - \cfrac{1}{2}\Big) = \Big(7, 2\cfrac{3}{4}\Big))
does lie on the tangent at
(evident since that tangent has gradient -1 and so a
requires
, making the shift from
to
:
as required
- We immediately know the coordinates of
are
because
is the midpoint of
, but this could also be established by a similar method using the two normals
- We have confirmed that
is a square as
(as
is horizontal and
is vertical)
- We have the area as half the square of the diagonal, or we could show that all four sides have the same length
and then say that the area is ^2 = \cfrac{1}{2})