Solve 3 Points on a Line: Point C 0.5 AC from A to B

[onako]

Given points A(x1, y1) and B(x2, y2), I'm supposed to derive the coordinate of the point C
which should be on the line AB (such that B is on the line segment AC), such that the distance CB is 0.5*distance AC. The point is to avoid the solution where C is on "the other side", such that A is between C and B. We should obtain the solution where B is between A and C.

Thanks

 

[tiny-tim]

Welcome to PF!

Hi onako! Welcome to PF!

(try using the X₂ tag just above the Reply box )

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

 

[onako]

I know the coordinates when point B is the midpoint of AC (2*x2-x1, 2*y2-y1), but here the distance BC is 0.5*distance AB. I would need coordinates expressed with this percent variable (0.5 here). Thanks

 

[tiny-tim]

If O is the origin (0,0),

then express AB in terms of OA and OB.

 

[onako]

If O is the origin (0,0),

then express AB in terms of OA and OB.

I don't think this would lead to the solution.
Any other thoughts?

 

[Simon43254]

I agree with what Tiny-tim said, effectively he's trying to get you to think of this as the vector equation of a line and work it out from there. Rather than using the kinda of standard co-ordinate reasoning solution. The way I would do this is to draw a graph and using OA and OB, think of how you could find C from this.

 

[magpies]

The way the question is worded is probably where the problem is it's not really that clear imo.

 

[Martin Rattigan]

I know the coordinates when point B is the midpoint of AC (2*x2-x1, 2*y2-y1), but here the distance BC is 0.5*distance AB. I would need coordinates expressed with this percent variable (0.5 here). Thanks

Points on AB are given by (x_1+\lambda(x_2-x_1),y_1+\lambda(y_2-y_1)) where \lambda is the ratio of directed segments \overrightarrow{AC}/\overrightarrow{AB}. When \lambda=2 that gives the point you quoted with \overrightarrow{AC}/\overrightarrow{AB}=2.

Originally you said you wanted, "the solution where B is between A and C", with, "distance CB is 0.5*distanceAC". The sense of the directed segment \overrightarrow{CB} must be opposite from that of \overrightarrow{AC} to meet the first requirement, so \overrightarrow{AC}=2\overrightarrow{BC}, whence \overrightarrow{AB}=\overrightarrow{BC} and \overrightarrow{AC}/\overrightarrow{AB}=2. The point you gave is therefore the point C you originally requested, and, yes B is the midpoint of AC.

When you say in the above quote, "BC is 0.5*distance AB", this is not what you originally asked for, but if you want a formula based on \overrightarrow{BC}/\overrightarrow{AB}=-\overrightarrow{BC}/\overrightarrow{BA} then you can get that just by switching the A and B coördinates in the above formula.

 

[SonyAD]

Given points A(x1, y1) and B(x2, y2), I'm supposed to derive the coordinate of the point C
which should be on the line AB (such that B is on the line segment AC), such that the distance CB is 0.5*distance AC. The point is to avoid the solution where C is on "the other side", such that A is between C and B. We should obtain the solution where B is between A and C.

Thanks

Ok. Let's simplify the problem.

Let A(x₁), B(x₂) and C(x₃) be three points on a line.

Let us define the aliases:

a for x₁
b for x₂
c for x₃

Let us define the function hmttr(a,b) as

\frac{|b - a| + b - a}{2}

hmttr(a,b) returns b - a if b >= a and 0 otherwise.

Now we can define c as:

b + \frac{ hmttr(a,b) }{2} - \frac{ hmttr(b,a) }{2}

That is, b + half of how much to the right b is from a - half of how much to the right a is from b.

Now we have:

b + \frac{ |b - a| + b - a }{4} - \frac{ |a - b| + a - b }{4}

which works out to

\frac{3b - a}{2}

or

x_{3} = \frac{3x_{2} - x_{1}}{2}

This will work regardless whether B is to the right of A or vice versa. C will always wind up on the opposite side of B from A.

Now you solve the line equation for this result to find y₃ and z₃.

The abs() function is the mother of logic. Using it you can implement decisional blocks in hardware without logic circuits. I have used it to compute the sign of a number.