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A------B---------------C

A and C are moving points that can move anywhere

A = (xa, ya), C = (xc, yc)

B (xb, yb) is a point between A and C colinearly

With one condition that distance of AB = constant = r, distance of AC can be varied.

What is the value of xb and yb?

EDIT: Attempts done:

  1. Gradient

$m=\frac{x_{c}-x_{b}}{y_{c}-y_{b}}=\frac{x_{b}-x_{a}}{y_{b}-y_{a}}$

Problem: x and y mixed together.

  1. Distance Formula

$AC = AB + BC = \sqrt{(x_{c}+x_{a})^{2}+(y_{c}+y_{a})^{2}}$

Problem: x and y mixed together.

EDIT 2:

$(x-x_{a})^{2}+(y-y_{a})^{2}=AB \dashrightarrow$ Equation 1

Gradient of AC$=\frac{y_c-y_a}{x_c-x_a}=\frac{y-y_a}{x-x_a}$

$y-y_a=\frac{y_c-y_a}{x_c-x_a}(x-x_a)\dashrightarrow$ Equation 2

Subsitude Eq 2 into Eq 1,

$x=\sqrt{\frac{AB}{1+\frac{y_c-y_a}{x_c-x_a}}}+x_a$

Subsitude x into Eq 2,

$y=\frac{y_c-y_a}{x_c-x_a}(\sqrt{\frac{AB}{1+\frac{y_c-y_a}{x_c-x_a}}})+y_a$

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  • $\begingroup$ Please, show your attempt? $\endgroup$ – ً ً Jun 27 '16 at 6:47
  • $\begingroup$ Being too long not touching Maths, need to do a pattern which A and C will be moved and B follows. Example: mathopenref.com/collinear.html $\endgroup$ – Fatty Mieo Jun 27 '16 at 7:00
  • $\begingroup$ @FattyMieo Remove that heading please, it will only encourage downvoting. $\endgroup$ – A---B Jun 27 '16 at 9:52
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Hint: the locus of all possible points $B$ is a circle centred on $A$ and of radius $r$. The intersection between this circle and line $AC$ will give you point $B$ (and another point opposite it on the circle which you can neglect).

Can you write down the relevant equations and solve them?

EDIT: Sorry for the delay, I have been sick.

More complete solution.

Locus of possible points $B$: $\displaystyle (x-x_a)^2 + (y-y_a)^2 = r^2$.

Equation of line $AC$: $y = \displaystyle (x-x_a)(\frac{y_c - y_a}{x_c - x_a}) + y_a$

Solving for the intersection,

$$ (x-x_a)^2 + (x-x_a)^2(\frac{y_c - y_a}{x_c - x_a})^2 = r^2$$

$$(x-x_a)^2(\frac{(x_c - x_a)^2 + (y_c - y_a)^2}{(x_c - x_a)^2}) = r^2$$

$$x = x_a \pm \frac{r(x_c - x_a)}{\sqrt{(x_c - x_a)^2 + (y_c - y_a)^2}}$$

and

$$y = y_a \pm \frac{r(y_c - y_a)}{\sqrt{(x_c - x_a)^2 + (y_c - y_a)^2}}$$

That gives you two possible coordinates for $B$, namely:

$$(x_a + \frac{r(x_c - x_a)}{\sqrt{(x_c - x_a)^2 + (y_c - y_a)^2}}, y_a + \frac{r(y_c - y_a)}{\sqrt{(x_c - x_a)^2 + (y_c - y_a)^2}})$$

and

$$(x_a - \frac{r(x_c - x_a)}{\sqrt{(x_c - x_a)^2 + (y_c - y_a)^2}}, y_a - \frac{r(y_c - y_a)}{\sqrt{(x_c - x_a)^2 + (y_c - y_a)^2}})$$

One of these represents the point you require between $A$ and $C$, the other represents the antipodal point lying outside the segment $AC$. You can distinguish between these using distance arguments, but that is tedious.

Better is to consider the dot product. For a point $B$ between $A$ and $C$, you would expect $\vec{AB} \cdot \vec{AC} > 0$. But for a point $B$ outside the segment $AC$, you would expect $\vec{AB}\cdot \vec{AC} < 0$.

$$\vec{AB} \cdot \vec{AC} = \begin{bmatrix}x-x_a \\ y-y_a\end{bmatrix} \cdot \begin{bmatrix}x_c-x_a \\ y_c-y_a\end{bmatrix} = (x-x_a)(x_c - x_a) + (y - y_a)(y_c - y_a) = \pm \frac{r(x_c - x_a)^2}{\sqrt{(x_c - x_a)^2 + (y_c - y_a)^2}} \pm \frac{r(y_c - y_a)^2}{\sqrt{(x_c - x_a)^2 + (y_c - y_a)^2}} $$

which is clearly positive only for the first set of coordinates. So that's the solution you require.

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  • $\begingroup$ Edit 2, like that? $\endgroup$ – Fatty Mieo Jun 27 '16 at 7:45
  • $\begingroup$ @FattyMieo Use section formula. $\endgroup$ – A---B Jun 27 '16 at 9:55
  • $\begingroup$ AB is fixed, AC can be any length, a section formula won't fulfill :( $\endgroup$ – Fatty Mieo Jun 27 '16 at 10:07
  • $\begingroup$ @FattyMieo Sorry, I've been ill. I've just posted a detailed solution. $\endgroup$ – Deepak Jun 28 '16 at 2:17
  • $\begingroup$ Thank you, I hope you recover soon. :) $\endgroup$ – Fatty Mieo Jun 29 '16 at 5:47
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I don't think that the slope-like expressions such as $\frac{x_a-x_c}{y_a-y_c}$ will be particularly helpful in this problem.

Instead, this is most easily expressed in vector notation: $$B = A + \frac{r}{|C-A|} (C-A) $$ To make it more explicit, it can be expanded out: $$(x_b,y_b) = (x_a,y_a) + \frac{r}{\sqrt{(x_c-x_a)^2 + (y_c-y_a)^2}} (x_c-x_a,y_c-y_a) $$

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