Distinct Points

Introduction for distinct point:

The distance between any two different points (x1, y1) and (x2, y2).  The distance between two different points is basic concept in geometry. We now give an algebraic expression for the same.  Let P1 (x1, y1) and P2(x2, y2) be two distinct points in the Cartesian plane and denote the distance between P1 and P2 by d(P1, P2) or by P1P2. Draw the line segment P1P2. There are three cases are following.

Cases for Distinct Point

Case (i):

The segment `bar (P_(1)P_(2))` is parallel to the x-axis.  Then y1 = y2. Illustrate P1L and P2M, perpendicular in the direction of the y-axis. Then d(P1, P2) is equal to the distance between L and M.  But L is (x1, 0) and M is (x2, 0). So the length LM = |x1 – x2|.  Hence d(P1, P2) = |x1 – x2|.

Case (ii):

The segment `bar (P_(1)P_(2))` is parallel to the y-axis.  Then x1 = x2. Illustrate P1L and P2M, perpendicular in the direction of the y-axis. Then d(P1, P2) is equal to the distance between L and M.  But L is (0, y1) and M is (0, y2). So the length LM = |y1 – y2|.  Hence d(P1, P2) = |y1 – y2|.

Case (iii):

The line segment `bar (P_(1)P_(2))` is neither parallel to the x-axis nor parallel to the y-axis. Draw a line through P1 parallel to x-axis and a line through P2  parallel to y-axis. Let these lines intersect at the point P3. Then P3 (x2, y1). The length of the line segment P1P3 is |x1-x2| and the length of the segment P3P2 is |y1-y2|. We observe that the triangle ΔP1P3P2 is a right triangle.

Formula for distinct point:

`sqrt((x_(2) - x_(1)^(2)) + (y_(2) - y_(1))^(2))`

Problems for Distinct Points:

Let us some problems of distinct points:

Problem 1:

Find the distance between the points A(10, 5) and B(4, 8).

Solution:

Let d is the distance between the two points A and B.

Formula for distinct point:

`sqrt((x_(2) - x_(1))^(2)) + (y_(2) - y_(1))^(2))`

` = sqrt((4 - 10^(2)) + (8 - 5)^(2))`

`= sqrt( ((-6)^(2)) + (3)^(2))`

`= sqrt (36 + 9)`

` =sqrt ( 45)`

` = 3sqrt ( 5)`

So, the dietance is `3sqrt(5)`

Problem 2:

Find the distance between the points A(7, 11) and B(20, 10).

Solution:

Let d is the distance between the two points A and B.

Formula for distinct point:

`sqrt((x_(2) - x_(1)^(2)) + (y_(2) - y_(1))^(2))`

`= sqrt((11 - 7^(2)) + (20 - 10)^(2))`

`= sqrt( ((4)^(2)) + (10)^(2))`

`= sqrt ( 160)`

` = 4sqrt ( 10)`

`These are problems of distinct points.`

Least-squares Line

Introduction to least squares line:

There are many methods avilable for curve fitting. The most popular method of curve fitting is the principle of least squares line method. Curve fitting is a process of finding a functional relationship betweent the variables. It is useful in the study of correlation and regression.

Definition of least Squares Line :

Let (xi, yi) be the observed set of values of the variables (x, y), where i = 1, 2, 3,…,n. Let y = f(x) be  a functional relationship between x and y. Then di = yi - f(xi) which is the difference between the observed value of y and the value of y is determined by the functional relation is called the residuals. The priniciple of least squares states that the parameters involved in f(x) should be chosen in such a way that `sum`  di2 is minimum.

Fitting a Straight Line Using least Squares Method

Consider the fitting of the straight line y = ax + b to the data (xi, yi), i = 1, 2, 3, …, n. The residual

di is given by di = yi - (axi + b).

Therefore, `sum` di2 =`sum`  (yi - axi - b)2 = R (say).

Since we are using the principle of least squares, we have to determine the value of a and b so that R is minimum.

Determine the Parameters of a and B Using Leats Squares Line Method:

Since R is minimum, `(del R)/(dela)`   = 0 `=>` - 2 `sum` (yi - axi - b)xi = 0

`=>`  `sum` (xiyi - axi2 - bxi) = 0.

Therefore, a`sum`xi2 + b `sum`xi =  `sum`xiyi  ————(1)

`(del R)/(del b)`   = 0 `=>` - 2 `sum` (yi - axi - b) = 0

Therefore, a`sum`xi + nb =  `sum`yi  ————(2)

Equations (1) and (2)  are called normal equations from which a and b can be found.

Note:  If the given data is not in linear form it can be brought to linear form by some suitable transformation of variables. Then using the priniciple of least squares the curve of best fit can be achieved.

Polar Coordinates R

Introduction :

The polar coordinates R system is an option for rectangular system. In polar coordinate system, instead of a using (x, y) coordinates, a point is represented by (r, θ). Where r delineate the length of a straight line from the point to the origin and θ delineate the angle that straight line makes with the horizontal axis. The θ as the angular coordinate and r is generally referred to as the radial coordinate. From the origin the distance of a point P is consider by a point r (an arbitrary fixed point provided by the symbol Q).

Equations for Polar Coordinates R:

Consider θ =angle between the radial line from point P to Q and the given line “θ = 0”, a kind of positive axis for polar coordinates r system. Polar coordinates r are referred in terms of ordinary Cartesian coordinates through the transformations

x = r cos θ
y = r sin θ

Where r ≥0 0≤ θ < 2π.

From these relation we can see that the polar coordinates r of point P delineates the Relation x2 + y2 = r2 (cos2 θ + sin2 θ) ⇒ x2 + y2 = r2 (so that, as we indicated, P(x, y) point is on a circle of radius r centered at Q), other hand, we can find θ by calculating the equation

tan θ = y/x =⇒ θ = arctan (y/x),

for θ in the interval 0 ≤ θ < 2π.

Examples of Polar Coordinates R:

1) The following are typical “slices” in polar coordinates r (see the margin):

Radial slice = {(r, θ): θ = π/4, 1 ≤ r ≤ 2}

Radial slice = {(r, θ): θ = 3π/2, 0.5 ≤ r ≤ 0.8}

Circular slice = {(r, θ): r = 1.2, π/4≤ θ ≤ π/2}

Circular slice = {(r, θ): r = 3, 3π/4≤ θ ≤ π}

Now we can start describing regions using slices.

2) The ideas in Example 6 show that the circumference, C, of the circle x2 + y2 = R2 can be described by both in polar coordinates r.

C = {(r, θ): r = R, and 0 ≤ θ < 2π},

Along with the Cartesian description

C = {(x, y): |y| = R2 − x2, and − R ≤ x ≤ R}.