How can I calculate Distance of line? 
I have a picture. I want to measure of $P_{1}$ to $P_{4}$ distance.
Also I know $P_{1}$ to $P_{2}$ to real distance. and $P_{2}$ to $P_{3}$ real distance
$P_{1}-P_{2}$ real distance = $100$ mm ( $P_{1}-P_{2}$ picture size = $43,41$ ) and $P_{2}-P_{3}$ real distance = $100$ mm ( $P_{2}-P_{3}$ picture distance $31,45$ ) As you see. There is perspective. So on image distance are different.
How can I formula it. So how can I calculate $P_{1}$ to $P_{4}$ real distance to use picture distance.
So sory my bad English. I hope I mentioned about my question clearly.
 A: Ignoring possible aberrations, a picture can be modelled as a central projection on a plane of a 3D space.

In the figure is e represented the more simple situation, that is when the plane of projection is orthogonal to the horizontal plane. The point of projection is the eye of the observer $O$, whose position is defined by the distance from the plane of projection $d$ and its height $h$ on the horizontal plane.
The points $a_i$ are on a central straight line $r_0$, i.e. the line of intersection of the horizontal plane and the plane orthogonal to the projection plane and passing through the observer. The points $b_i$ are the projections of $a_i$ and we can see, from similarity that: $b_i :a_i=h:(d+a_i)$ so that:
$$
b_i=\dfrac{ha_i}{d+a_i}
$$
note that if $a_i \rightarrow \infty$ than $b_i \rightarrow h$: the ''point at infinity'' of the perspective.
Now consider a straight line $r_1$ on the horizontal plane and parallel to $r_0$ and points $c i \in r_1$ that are orthogonal projections of $a_i$. 

The second figure represents this situation in the projection plane where $d_i$ are the projections of $c_i$ and also here you have similar triangles that permits to calculate the position (and distance) of the points $d_i$ starting from the $c_i$.
In your case you can find the parameters $d$ and $h$ using the know measures for $P_1,P_2,P_3$, than find the real distance of $P_4$ using the apparent distance measured on the photo.
This is the simpler situation, when the plane of projection is vertical. If this plane is inclined , the procedure is similar. 
A: Thank you vey much yor answer. Also I find to solution my problem.  I use cross ratio.
As your first picture;
| $a_1$$a_3$| * |$a_2$$a_4$| / |$a_2$$a_3$| * |$a_1$$a_4$| =|$b_1$$b_3$*| * |$b_2$$b_4$| / |$b_2$$b_3$| * |$b_1$$b_4$|
a distanses are real distanse. b distances are picture distance.
