Discovering a fourth mysterious point.

[TheAgustin]

Hi there. As usual begging for your help in math issues

This is the situation, I've three (3) points in a 2-d area (A , B and C)
I know the coordinates of these points.

From there I need to discover the coordinates of a fourth point (D) which is related to the previous three points.

If you look at the attachment you will understand it easily.

The fourth point lies in segment A-B in such a place where it creates two straight angles when originating the segment D-C

Is there a simple formula I can use? I'm familiar with formula for calculating distance.

Thanks a lot for your precious time and efforts.
Have a nice day.
Agustín

[Logophobic]

Measure either angle ABC or BAC and calculate the distance to C from the vertex. You can then calculate the distance to D.

[TheAgustin]

Logophobic, it's a little unclear answer for me. Could you develop a little further to see if this could help in finding point D coordinates (x,y).

Thanks for your interest!

[kberry79]

If AB and CD are perpendicular, then you can use the coordinates to find the equations of the lines that pass through AB and CD. Then find the point of intersection of the two lines.

[TheAgustin]

Dear KBerry79,

Yes, A-B and C-D are perpendicular to each other, that's what I know.
I also know the coordinates of points A, B and C and IGNORE the coordinates of D.
Coordinates of point D may vary depending on the coordinates of points A, B and C.

I need an ecuation to calculate d1 and d2 based on the things I know (a1, a2, b1, b2, c1, c2, AB CD perpendicular)

[kberry79]

The equation of the line through AB is
(y - a2) = (a2 - a1)/(b2 - b1) * (x - a1).

The equation of the line through CD is
(y - c2) = -(b2 - b1)/(a2 - a1) * (x - c1).

Solve both equations for y. Set the two expressions of x equal to each other and solve for x. This is your x-coordinate of D. Plug it into one of the other to equations and solve for y. This is the y-coordinate of D.

[Logophobic]

kberry79's equations are incorrect, but the approach is probably best.

The equation for line through AB is
(y - a2) = (b2 - a2)/(b1 - a1) * (x - a1)

The equation for line through CD is
(y - c2) = -(b1 - a1)/(b2 - a2) * (x - c1)

[kberry79]

oops...sorry about the confusion...Logophobic is right
dy = b2 - a2 not a2 - a1, etc. ... lol

[Mattywoo2]

Just use vectors.

Firstly, call the origin O. Then say OA=a, OB=b, OC=c and OD=d.

The line AB has equation a + mu(b-a), where mu is a scalar.

The line DC has equation d + lambda(c-d), where lambda is a scalar.

The dot product of the two direction vectors [i.e: (b-a).(c-d)] will equal zero, since they are perpendicular. By evaluating the dot product you'll end up with an expression with d1 and d2 in it. It might look something like 2d1 + d2 = 3, but of course it will entirely depend on the values of a1, a2, b1, and b2.

Now, because d lies on the line AB, we can say d = a + mu(b-a) for some value of mu.

If you compare the i and j parts on both sides of the equation, you'll get two equations. One will be like d1 = "something with i's in" the other will be like d2 = "something with j's in". Both will have mu in them. Now if you use the equation you found when you did the dot product, you can solve for mu.

Once you have the mu value, substitute it back into the equation d = a + mu(b-a). That will give you your d.


If any of this is unclear, give me the values of a1, a2, b1, b2, c1 and c2 and I'll do the solution for you as a nice pdf. Good luck.