MHB Prove Triangle Inscribed in Semicircle Is Right Angle | Arundev Answers

AI Thread Summary
The discussion focuses on proving that an angle inscribed in a semicircle is a right angle using coordinate geometry. A unit semicircle centered at the origin is considered, with point P defined as (x, √(1-x²)). The slopes of the line segments from point P to the endpoints of the semicircle are calculated, showing that the product of these slopes equals -1. This confirms that the lines are perpendicular, thereby proving that the triangle formed is a right triangle. The conclusion reinforces the geometric principle that an angle inscribed in a semicircle is indeed a right angle.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

Using coordinate geometry prove that angle in a semicircle is a right angle?

I have posted a link there to this thread so the OP can view my work.
 
Mathematics news on Phys.org
Hello Arundev,

Consider the following diagram:

https://www.physicsforums.com/attachments/1763._xfImport

Without loss of generality, I have chosen a unit semicircle whose center is at the origin.

Point $P$ is $$(x,y)=\left(x,\sqrt{1-x^2} \right)$$.

The slope of line segment $A$ is:

$$m_A=\frac{\sqrt{1-x^2}-0}{x-(-1)}=\frac{\sqrt{1-x^2}}{1+x}=\sqrt{\frac{1-x}{1+x}}$$

The slope of line segment $B$ is:

$$m_B=\frac{\sqrt{1-x^2}-0}{x-1}=-\frac{\sqrt{1-x^2}}{1-x}=-\sqrt{\frac{1+x}{1-x}}$$

As proven http://mathhelpboards.com/math-notes-49/perpendicular-lines-product-their-slopes-2953.html, two lines are perpendicular if the prodict of their slopes is $-1$.

$$m_Am_B=\left(\sqrt{\frac{1-x}{1+x}} \right)\left(-\sqrt{\frac{1+x}{1-x}} \right)=-1$$

Thus, we know line segments $A$ and $B$ are perpendicular, and so the triangle is a right triangle.
 

Attachments

  • arundev.jpg
    arundev.jpg
    6.8 KB · Views: 135
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top