[Math] Given 2 Points of a Rectangle, find the Center

geometry

In a feature that I am writing for a program, I need to find the (x, y) of the center of a rectangle while only being provided the (x, y) of the top left and bottom right corners. I've been out of a math course for awhile now and cannot thing of any equations that I can use to solve this. Here is a Paint image I drew up to show what I am working with. Note that the points are going to be dynamic, so the numbers won't ever be the same. If someone can point me in the right direction, that would be great!

Best Answer

It's just the midpoint between the two corners. If the points are $(p_1,p_2)$ and $(q_1,q_2)$, then the midpoint is $$\left(\tfrac{p_1+q_1}2,\tfrac{p_2+q_2}2\right)$$ the coordinate-wise "average" of the two points.

The final result:

$$\LARGE\boxed{R=\frac H2+\frac{W^2}{8H}}$$

float cos_t = (R-H)/R;
P_x=(rect.width/2);
P_y=-R.cos_t;

Process:

enter image description here

Without using calculus(actually it would never be solved with calculus) and only using trigonometry:

The component of R along $\theta$ is $R.\cos\theta$ and perpendicular to it is $R.\sin\theta$

Now remaining component is $R-Rcos\theta=R(1-\cos\theta)$

Now we have: $$ W=2.R.\sin\theta\implies\sin\theta=\frac W{2R}\\ H=R(1-\cos\theta)\implies\cos\theta=1-\frac HR=\frac{R-H}R $$ Using the identity $\sin^2\theta+\cos^2\theta=1$ $$\left(\frac W{2R}\right)^2+\left(\frac{R-H}R\right)^2=1$$ $$\frac{W^2}{4R^2}+\frac{R^2+H^2-2RH}{R^2}=1\\\text{ I assume you are familiar with }(a\pm b)^2=a^2+b^2\pm 2ab$$ $$\frac{W^2}4+R^2+H^2-2RH=R^2$$ $$\frac{W^2}4+H^2=2RH$$ $$R=\frac{W^2}{8H}+\frac H2$$ It is easy to see that center is at the midpoint of width. Also y-coordinate is $-R.\cos\theta$

Best Answer

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The final result:

Process:

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