How to move a point along a line

Question

Help me understand intuitively how one might move a point along a line. I'm not able to follow various solutions. These solutions detail steps to follow but I'm not understanding why these steps are laid out/suggested. (It may be a good disclaimer that my math education taught me to memorize theorems and not how to apply them).

Help me work from where I am at to what I need to understand to move a point along a line.

1. We have 2 points, A and B. A is at (0, 2) and B is at (4, 1).
2. Building from the Pythagorean theorem, the distance between points A and B is $\sqrt{(1-2)^2 + (4-0)^2}$.
3. Simplifying results in $\sqrt{(-1)^2 + (4)^2}$.
4. Simplifying results in $\sqrt{1 + 16}$.
5. Simplifying results in $\sqrt{17}$.
6. Simplifying results in ~4.12.

I can get this far, but then I do not know where to go.

Let's say I want to move 2 units away from point A, I could approach it like this.

1. Since the distance from A to B is ~4, I should find an x-value halfway between A and B; I choose 2. $(0 + 4) / 2$.
2. In order to find the y-value at a given x, I need to know the slope of the line, so I remember that the slope is $(y2 - y1) / (x2 - x1)$.
3. Simplifying results in $(1 - 2) / (4 - 0)$.
4. Simplifying results in $-1 / 4$.
5. The slope of the line is $-\frac{1}{4}$.

I can now find the y-value of the line at an x-value of 2.

1. Solve $y = mx + b$.
2. Simplifying results in $y = -\frac{1}{4}*2 + 2$.
3. Simplifying results in $y = -\frac{1}{2} + 2$.
4. Simplifying results in $y = -\frac{1}{2} + \frac{4}{2}$.
5. Simplifying results in $y = \frac{3}{2}$ = $1.5$.

This approach is naive, but it's all I can come up on my own. It also doesn't work if the line is vertical because there is no midpoint x-value we can use.

Answer 1

If you move $4$ units horizontally, you move $1$ unit vertically, and $\sqrt{17}$ units obliquely.

Now everything is proportional, and

if you move $\dfrac4{\sqrt{17}}$ units horizontally, you move $\dfrac1{\sqrt{17}}$ unit vertically, and $1$ unit obliquely.

Answer 2

The locus of points at constant distance from a given point is a circle, with the equation $(x-x_c)^2+(y-y_c)^2=r^2$, with subscript $c$ denoting the value at the centre (your given, starting point) and $r$ the radius of the circle (your desired distance).

Now you want the points on this circle which are also on your line - this is simply their intersection.

You can then determine the equation of the line, use your preferred method for solving the resulting set of simultaneous equations (if using the line $y=mx+c$, substitution of $y$ followed by simplification and solution of the quadratic is straightforward) giving you two values for the $x$-ordinate of your new point.

Use the equations to find the $y$-values, then use your knowledge of the relative locations of the lines, and any other information about how you are moving to your new point, to choose the correct one.

Answer 3

You may use the intercept theorem, https://en.wikipedia.org/wiki/Intercept_theorem From this you get $$\frac{\sqrt{17}-2}{\sqrt{17}}=\frac{4-x}{4},$$ hence $x=8/\sqrt{17}$ which then may be substituted in the equation of the line through $A$ and $B$ or use the theorem again. In either case you get $y=2-4/\sqrt{17}$.

Answer 4

If $l=\overline{AB}$, then the coordinates of a point $P$ at a distance $|t|$ from $A$, on the same side as $B$ for $t>0$ and on the other side if $t<0$, can be written as: $$ P=A+{t\over l}(B-A), $$ where the sum or difference of two points is computed coordinate-wise, as you do with vectors.

In your example: $B=(0,2)$, $A=(4,1)$, $l=\sqrt{17}$, one gets: $$ P=(4,1)+{t\over\sqrt{17}}[(0,2)-(4,1)]= (4,1)+{t\over\sqrt{17}}(-4,1)= \left(4-4{t\over\sqrt{17}},1+{t\over\sqrt{17}}\right). $$ A geometric proof of that relies on similar triangles and is not difficult. If you are familiar with vectors, you can observe that $(B-A)/l$ is a unit vector in the direction $\vec{AB}$.

EDIT.

Sketch of proof (see figure below). $$ {PK\over PA}={BH\over BA}\implies {y_P-y_A\over t}={y_B-y_A\over l}\implies y_P=y_A+{t\over l}(y_B-y_A), $$ and the analogous for $x$ coordinates. Of course you should check this works for all possible relative positions of $A$, $B$ and $P$.