Finding a missing co-ordinate on a right-angle triangle

[maye]

Hi

Could someone tell me please, how do you find a missing co-ordinate (on a right-angle triangle)?

Example:

You have points A, B and C, with co-ordinates (2, 1), (b, 3) and (5, 5).
We know that b > 3 and angle ABC = 90degs.

Which method is the simplest to use in problems such as this?

Thanks in advance!

 

[HallsofIvy]

Pythagorean theorem: \(\displaystyle a^2+ b^2= c^2\). When you see "right angle triangle" that should be the first thing that comes to mind.
A= (2, 1), B= (b, 3), and C= (5, 5). Since the right angle is at B, one leg has length \(\displaystyle \sqrt{(2- b)^2+ (1- 3)^2}= \sqrt{b^2- 4b+ 8}\), the other has length \(\displaystyle \sqrt{(5- b)^2+ (3- 5)^2}= \sqrt{b^2- 10b+ 29}\), and the hypotenuse has length \(\displaystyle \sqrt{(2- 5)^2+ (1- 5)^2}= \sqrt{9+ 16}= \sqrt{25}= 5\).

So \(\displaystyle (b^2- 4b+ 8)+ (b^2- 10b+ 29)= 25\).

 

[maye]

Pythagorean theorem: \(\displaystyle a^3+ b^2= c^2\). When you see "right angle triangle" that should be the first thing that comes to mind.
A= (2, 1), B= (b, 3), and C= (5, 5). Since the right angle is at B, one leg has length \(\displaystyle \sqrt{(2- b)^2+ (1- 3)^2}= \sqrt{b^2- 4b+ 8}\), the other has length \(\displaystyle \sqrt{(5- b)^2+ (3- 5)^2}= \sqrt{b^2- 10b+ 29}\), and the hypotenuse has length \(\displaystyle \sqrt{(2- 5)^2+ (1- 5)^2}= \sqrt{9+ 16}= \sqrt{25}= 5\).

So \(\displaystyle (b^2- 4b+ 8)+ (b^2- 10b+ 29)= 25\).

Thanks a lot for replying.

I understand that you use pythagoras' theorem for finding length but what about if you want to just find what co-ordinate 'b' is?

Thanks!

 

[Deleted member 4993]

Thanks a lot for replying.

I understand that you use pythagoras' theorem for finding length but what about if you want to just find what co-ordinate 'b' is?

Thanks!

HoI derived an equation for 'b' → [FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]b[/FONT]^{[FONT=MathJax_Main]2[/FONT]}[FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math-italic]b[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Main]) [/FONT][FONT=MathJax_Main]+ [/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]b[/FONT]^{[FONT=MathJax_Main]2[/FONT]}[FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]10[/FONT][FONT=MathJax_Math-italic]b[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]29[/FONT][FONT=MathJax_Main]) [/FONT][FONT=MathJax_Main]= [/FONT][FONT=MathJax_Main]25[/FONT] → from which you should be able to calculate 'b'!