Given 𝑃1 and 𝑃2, how do you calculate 𝑃3 and 𝑃4 after shortening the line segment by equal distance on each point

geometrylinear algebratrigonometry

Let's say I have a line segment given 𝑃1 and 𝑃2. I can calculate the Distance between the two, and the Slope, which will help.

I want to 'shorten' the line segment by equal Distance X on both points of the line segment. Therefore, 𝑃1 and 𝑃2 will move X distance closer to each other producing new points 𝑃3 and 𝑃4. Together these new points will be a 'shortened' line segment, with the same exact slope and intermediate points between the 𝑃3 and 𝑃4.

How Do I calculate 𝑃3 and 𝑃4? (assuming both are the same process)

Best Answer

Suppose distance between points $P,Q$ be denoted by $d(P,Q)$. Then you can find $d(P_1,P_2)$, call it $D$.

$P_3$ divides the segment $\overline{P_1P_2}$ internally in the ratio $X:(D-X)$, so you can use the section formula $$ P_3 = \dfrac{(D-X)P_1+XP_2}D \tag{1}$$

and $P_4$ divides the segment $\overline{P_1P_2}$ internally in the ratio $(D-X):X$, so $$ P_4 = \dfrac{XP_1+(D-X)P_2}D$$

Note: If $P$ has coordinates $(x_P,y_P)$ (assuming you are working with $2$ coordinates, it works for any number of coordinates), then for $P_3=(x_{P_3},y_{P_3})$, $(1)$ means "apply it to each coordinate of $P_2$ and $P_3$", i.e. $$x_{P_3}=\dfrac{(D-X)\cdot x_{P_1}+X\cdot x_{P_2}}D\\ y_{P_3}=\dfrac{(D-X)\cdot y_{P_1}+X\cdot y_{P_2}}D$$

Related Solutions

[Math] Get a point with given distance to a line segment

Alright so here is one way of going about it (if I understood you correctly). So let $Z = (z_1,z_2)$ be the point lying on the line through $A,B$ such that the line passing through $C,Z$ is perpendicular to the line through $A,B$. You can find this point as follows.

We can find that $d(A,C) = \sqrt{104}$, $d(C,D) = \sqrt{(z_1-12)^2 + (z_2-10)^2}$, and $d(A,D) = \sqrt{(z_1-10)^2 + (z_2-20)^2}$.

So Pythagorean's theorem gives us the equation: $$(d(C,D))^2 + (d(A,D))^2 = 104.$$

But we also know that $(z_1,z_2)$ is a solution to the equation of the line through $A,B$, namely the line $y= -\frac{13}{25}x + \frac{630}{25}$.

So now you have two equations and two unknowns. Give that a try and see if you can find $Z$. Once you have $Z$ you should be able to figure out whatever else you were wondering about.

Given a line segment (AB) and point (C), find the furthest point between (AB) and (C) on the line segment with a max distance (D)

Thanks to Vasya, I realised that it just requires a circle around point C and then a function to find where the line intersects. Here's the code (C#):

static Vector2 IntersectCircle(Vector2 position, float radius, Vector2 start, Vector2 end) {
    var ax = start.X;
    var ay = start.Y;
    var bx = end.X;
    var by = end.Y;
    var cx = position.X;
    var cy = position.Y;
    var dx = bx - ax;
    var dy = by - ay;

    var a = dx * dx + dy * dy;
    if (a <= 0.000001f)
        return end;

    var b = 2 * (dx * (ax - cx) + dy * (ay - cy));
    var c = (ax - cx) * (ax - cx) + (ay - cy) * (ay - cy) - radius * radius;

    var t = (float)((-b - Math.Sqrt(b * b - 4 * a * c)) / (2 * a));
    return new Vector2(ax + t * dx, ay + t * dy);
}

Best Answer

Related Solutions

[Math] Get a point with given distance to a line segment

Given a line segment (AB) and point (C), find the furthest point between (AB) and (C) on the line segment with a max distance (D)

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