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How does a fractional distance calculator help compute the distance between two points in a fraction...

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A fractional distance calculator is a tool that computes the distance between two points based on a fractional or proportional value. This concept is particularly useful in fields such as regression analysis and statistical geometry, where it is often necessary to determine intermediate points or to measure distances that are not simply the direct Euclidean distance between two points.
A fractional distance calculator is a tool that computes the distance between two points based on a fractional or proportional value. This concept is particularly useful in fields such as regression analysis and statistical geometry, where it is often necessary to determine intermediate points or to measure distances that are not simply the direct Euclidean distance between two points.

Let's consider a scenario where we have two points in a two-dimensional space, A(x1,y1) and B(x2,y2), and we want to find a point C that lies on the line segment AB at a fractional distance from A. The fractional distance f is a value between 0 and 1 that represents the proportion of the distance from A to B at which point C is located.

The steps to calculate the coordinates of point C using a fractional distance f are as follows:

1. Determine the vector from A to B: Calculate the differences in the x and y coordinates of A and B.
Δx=x2−x1
Δy=y2−y1

2. Multiply the vector by the fractional distance f: This will give us the vector from A to C.
Δxf=f⋅Δx
Δyf=f⋅Δy

3. Calculate the coordinates of C: Add the vector from A to C to the coordinates of A.
xc=x1+Δxf
yc=y1+Δyf

Therefore, the coordinates of point C are given by:
C(xc,yc)=C(x1+f⋅Δx,y1+f⋅Δy)

In regression analysis, this concept can be applied to find points on a regression line. For example, if you have a linear regression model that predicts a dependent variable y based on an independent variable x, and you want to find the predicted value of y for a given x value that lies fractionally between two observed x values, you can use the fractional distance calculator to interpolate the corresponding y value.

In statistical geometry, fractional distances can be used to calculate centroids or centers of mass for complex shapes by integrating over the shape's area or volume with respect to a density function. This involves calculating the fractional distances of infinitesimally small elements and summing their contributions.

In summary, a fractional distance calculator is a versatile tool that can be used to find intermediate points between two known points in a proportional manner. This has applications in regression analysis for interpolation and in statistical geometry for calculating centroids or centers of mass.

Exercises 1-4 - Distance between two points in the coordinate plane

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This exercise has 4 parts,
all involving a particle moving from A to B in the plane.
In each case we're given the coordinates of A and B.
What we have to do is find 3 things: Delta x,
Delta y, and the distance from A to B.
Let's begin with Part 1.
Let's draw A and B on the coordinate plane below.
Delta x is the difference of the x's,
but the order matters.
We're going from A to B,
so let's draw the line
and put a little arrow here to show that we're going in this direction.
Delta x is equal
to the x(B) minus the x(A) minus 1 takeaway minus 3,
which is plus 2.
For Delta y, we need the differences in the Ys.
Once again, we're going from A to B.
We take the y(B) minus the y(A),
which is 0 minus 4,
so Delta y is minus 4.
The last thing we have to find is the distance from A to B.
Let's just join A and B.
There is a formula where d,
this is the distance d,
the formula is that d is equal to
the square root of Delta y^2 plus Delta x^2.
This is essentially an immediate application of
Pythagoras' theorem in a right-angle triangle.
In our case,
we get the square root of
Delta y^2 is minus 4^2.
Delta x^2 is 2^2.
Finally, this comes out to 16 plus 4 square root of 20.
We're done with Part 1.
Part 2 is very similar to Part 1,
just different numbers and I've pre-drawn
the coordinates A and B on the coordinate plane,
There's A minus 2,
2 and B minus 4, 2.
I've also drawn the triangle.
All we have to do is say what is Delta x,
Delta y, and the distance.
Delta x is the difference between the Xs,
but we have to take the x(B) minus the x(A),
which is minus 4,
less minus 2,
so that's minus 2.
Delta y is the y(B) minus the y(A),
2 minus minus 2,
which is 4,
and Delta x, Delta y.
Finally, we have to find out what the distances from A to B,
and we use the formula d equals the square root of Delta y^2 plus Delta x^2,
which is equal to the square root of minus 2^2.
That's what Delta x, Delta y is for plus 4^2,
which comes out to be square root of 20.
We're done with Part 2.
Exercise 3 is similar to the first two,
but there is a slight difference,
as we'll see when we plot the points on the coordinate plane,
we see that they run a horizontal line.
The A and the B have the same y-coordinate,
which means we can't really do a proper triangle here,
but we can still compute Delta x and Delta y. Delta x is the x(B) minus
the x which is minus 4.3 minus the x(A) minus minus 3.2,
the answer comes out to be minus 1.1.
Delta y is equal to the difference in the y-coordinates,
the y(B) minus the y(A),
which is minus 2, less minus 2,
which is 0, which is not surprising because there
is no vertical distance between A and B.
Finally, we compute the distance d using the same formula.
We had the d was equal to the square root of Delta y^2 plus Delta x^2,
which equals the square root of 0^2 plus minus 1.1^2.
The zero gives nothing,
this gives the square root of something squared is just
that something absolute value, which is 1.1.
We could've seen this right away because the horizontal distance Delta x is minus 1.1.
The distance is 1.1, but stick with the formula.
There's Delta x, there's Delta y,
and there's the distance, we're done with Part 3.
Exercise 4 is similar to the previous ones.
We have here a square root of 2 but since we're only approximating,
let's write the square root of 2 as 1.4.
That's close enough.
We'll sketch the points on the coordinate plane.
I've drawn the points as well as the triangle.
Now Delta x. Delta x
is equal to the difference in the x's but we have to take B minus A,
0 minus square root of 2,
which is minus the square root of 2.
Delta y. The difference in the y's but B minus A it's 2.5 minus 3,
which is minus 0.5.
The distance d, we can compute from the usual formula.
We have that d equals the square root
of Delta y^2 plus Delta x^2,
which equals the square root of Delta y^2 is
0.25 minus the square root
of 2^2 is just 2,
say we didn't need the approximation after all.
That's equal to the square root of 2.25.
That comes out to be a nice round number, almost 1.5.
We found Delta x, Delta y,
and d. We're done with Part 4 and with the whole exercise.
This video explains how to calculate the distance between two points on a coordinate plane. It covers four parts, each involving a particle moving from point A to point B. For each part, the video explains how to calculate Delta x, Delta y, and the distance from A to B. It also provides an immediate application of Pythagoras' theorem in a right-angle triangle.

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