How Do You Calculate Point P3 in a Coordinate System with a 90-Degree Angle?

[chuyler1]

See the attached image first. Given that I know points P1 and P2, what would the equation be to get point P3, given that the angle is 90 degrees and the distance from P2 to P3 is 1/5 the distance from P1 to P2?

Thanks in advance!

 

[tiny-tim]

Welcome to PF!

Hi chuyler1! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help.

 

[chuyler1]

I haven't a clue where to begin. It's been 8 years since the last time I had to think about calculus. I found some equations on how to rotate a point by 90 degrees but the only examples are to rotate around (0,0).

This calculation has to be performed hundreds of times in a short amount of time so if the equation is too complex i may just fudge it by increasing point P2's y by L/5. It simplifies the problem but doesn't provide desirable results in all situations.

 

[Redbelly98]

Could you use the fact that the slopes of perpendicular lines are negative reciprocals of each other? I.e.:

m1 = -1/m2

 

[chuyler1]

What is m1 and m2 in your equation?

 

[Redbelly98]

They are the slopes of the two lines you drew.

m1 = slope of line P1-P2
m2 = slope of line P2-P3

 

[robphy]

No calculus is needed here... just vector algebra.
(Will P1-P2 ever be vertical?)
Are you willing to use trig-functions?

 

[chuyler1]

Yes, P1 and P2 are arbitrary, they could be any angle or even reversed. I have trig functions available to me so a solution involving them isn't a problem.

 

[Redbelly98]

I've had some more time to think about this. We won't even need trig!

In the figure, note that we form two similar right triangles, with a scale factor of 1:5.

To get the x-coordinate of P3, subtract (y2-y1)/5 from x2.
To get the y-coordinate of P3, add (x2-x1)/5 to y2.

 

[chuyler1]

Awesome! Works perfectly!