In this post, I'll give a mathematical justification for the following statements about photos of line segments:

1: The photo of a straight line segment will be a straight line segment.

2: If the line segment is in a plane perpendicular to the direction that the camera is pointing, then the slope of the line in the photo will be the same as it is in reality.

As a result, if a camera is pointing in a horizontal direction, then vertical lines ( and vertical edges of objects ) will be vertical in the photo. Also a horizontal line which is in the plane perpendicular to the direction the camera is pointing will be horizontal in the photo.

I will show that those statements are true for regular photos, i.e. for photos which, when rendered as an image, maintain the original orientation, for an observer located in a particular location.

Consider a static image that is taken with a camera. We'll label the direction the camera is pointing as the \( x_3 \) axis. Suppose we render the image on a screen perpendicular to the \( x_3 \) asix, a distance \( \nu \) from the observer.
The image of a point will the be rendered on the screen precisely in the spot that is covering the original point.

With 3 dimensional coordinates, we'll specify the position of the observer to be at the origin. The screen is in the pane
\(x_3 = \nu \)

The \(x_1 \) axis is parallel to the ground and the \( x_2 \) axis is perpendicular to both \( x_1 \) and \( x_3 \).

When the camera is pointing in the horizontal plane, \( x_2 \) is vertical.

If the object is at \( \underline{x}^a = ( x_1^a , x_2^a, x_3^a ) \)

then the image should be somewhere along the line from the origin 0 to \( \underline{x}^a \)

In other words, for some real number \( \alpha \), the location of the images is given by:
\[ Image(\underline{x}^a) =\underline{x}^a \alpha = ( x_1^a , x_2^a , x_3^a ) \alpha \]
The image is on the screen so
\[ \nu = x_3^a \alpha \]
and so
\[ \alpha = \frac{\nu}{x_3^a} \]
hence
\[ Image(\underline{x}^a) = ( \frac{x_1^a}{x_3^a} , \frac{x_2^a}{x_3^a} , 1 ) \nu \]
Now suppose we had a line segment running from \( \underline{x}^a \) to \( \underline{x}^b \)

Let \( \underline{x}^{\lambda} \) be a point on that line segment such that:
\[ \underline{x}^{\lambda} = \underline{x}^a ( 1 - \lambda) + \underline{x}^b \lambda \]
where \( 0 \le \lambda \le 1\)

So
\[ Image(\underline{x}^{\lambda}) =
\left( \frac{x^a_1 (1 - \lambda ) + x^b_1 \lambda }{x^a_3 (1 - \lambda ) + x^b_3 \lambda} ,
\frac{x^a_2 (1 - \lambda ) + x^b_2 \lambda }{x^a_3 (1 - \lambda ) + x^b_3 \lambda} ,
1 \right) \nu \]
If we let
\[ \beta =\frac{ x^b_3 \lambda}{x^a_3 (1 - \lambda ) + x^b_3 \lambda} \]
then after doing some algebra we find:
\[ Image(\underline{x}^{\lambda}) =
\left( \frac{x^a_1}{x^a_3} ( 1 - \beta) + \frac{x^b_1}{x^b_3} \beta ,
\frac{x^a_2}{x^a_3} ( 1 - \beta) + \frac{x^b_2}{x^b_3} \beta ,
1 \right) \nu \]
And as we vary \( \lambda \), \( \beta \) changes and the image traces out a line segment on the screen from
\( ( \frac{x^a_1}{x^a_3} , \frac{x^a_2}{x^a_3}, 1 ) \nu \)
to
\( ( \frac{x^b_1}{x^b_3} , \frac{x^b_2}{x^b_3}, 1 ) \nu \)

So we have shown that in our model, the image of a straight line segment is a straight line segment.

It took some effort using algebra to get to that statement. It turns out we could have arrived at that statement using some geometry. The image of the line segment from \( \underline{x}^a\) to \( \underline{x}^b\) will be along the intersection of the plain \( x_3 = \nu \) and the plane containing \( \underline{x}^a\), \( \underline{x}^b\) and the origin. The intersection of two planes is a line and so the image of the straight line segment will be a straight line segment.

Again let us consider the image of a line segment from \( \underline{x}^a\) to \( \underline{x}^b\) but this time let's assume that both \( \underline{x}^a\) and \( \underline{x}^b\) are equidistant from the plane containing the screen at \( x_3 = \nu \).

So
\[ x^a_3 = x^b_3 \]
In this case, the image will be rendered along the line segment
from \( \left( \frac{x_1^a}{x^a_3}, \frac{x_2^a}{x^a_3} , 1 \right) \nu \)

to \( \left( \frac{x_1^b}{x^a_3}, \frac{x_2^b}{x^a_3} , 1 \right) \nu \)

Ignoring for a moment the \( x_3 \) axis, we find the slope of the line segment from \( \underline{x}^a\) to \( \underline{x}^b\), in the \( x_1 \) , \(x_2 \) plane is:
\[ slope_{12}( \underline{x}^a, \underline{x}^b) = \frac{x^b_2 - x^a_2}{x^b_1 - x^a_1} \]
which is the same as the slope of the line segment from \( Image( \underline{x}^a ) \)
to \( Image( \underline{x}^b ) \).

So in this case the slope is preserved.

From a practical point of view, if a camera is pointing in the horizontal plane and it is used to take a photo of a building with some vertical lines, then those vertical lines will be preserved in the photo.

The image of a horizontal line segment which is parallel to the \( x_3 = \nu \) pane ( parallel to the screen ), will be rendered as a horizontal line in the photo.

So if you take a photo with the camera pointing perpendicularly at a wall, then horizontal lines on that wall will be horizontal lines in the photo.

If you are perpendicular to a flat straight road, then a line at the edge of a road will appear as horizontal, going across a photo.

On the other hand, if the photographer is very close to a building and points the camera upwards, then the real vertical axis will not be parallel to the \( x_2 \) axis and so a real-life vertical line, may not appear to be vertical in the photo. In fact the building may appear to lean back in the photo.

All bets are off when a fish-eye lens is used and then the image is rendered on a flat screen. The objects in the image will not be aligned with the real objects and you may observe straight lines to appear to become curved in the photo.

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