Plug in and Solve Parabolas

Introduction to Plug in and Solve Parabolas:

A curve which is formed by the intersection of a right circular cone and half of the circle is called as parabola. Directrix of a parabola is a set of all points located at same distance from a fixed line. The fixed point is called as focus and not on the directrix. The midpoint between focus of a parabola and vertex of a parabola is called as vertex. A line passing through vertex and focus of a parabola is called as axis of symmetry. Finding the above four criteria by solving parabola equation. Let us see about plug in and solve parabolas in this article.

Worked Examples to Plug in and Solve Parabolas

The general form the parabolic curve is` y = ax^(2) + bx + c` or `y^(2) = 4ax` . Substitute the above formula to find the vertices, latus rectum, focus and axis of symmetry.

Example 1 for Plug in and Solve Parabolas – Vertex:

Find the vertex of a parabola equation `y = x^ (2) + 4x + 3` .

Solution:

Given parabola equation is `y = x^ (2) + 4x + 3` .

To find the vertices of a given parabola, we have to plug `y = 0` in the above equation, we get,

` 0 = x^ (2) + 4x + 3`

Now we have to factor the above equation, we get,

So `x^ (2) + x + 3x + 3 = 0`

`x(x + 1) + 3 (x + 1) = 0`

`(x + 1) (x + 3) = 0`

From this `x + 1 = 0` and `x + 3 = 0`

Then `x = - 1` and `x = - 3`

So, the vertices of given parabola equation is` (-1, 0)` and `(-3, 0)` .

Example 2 for Plug in and Solve Parabolas – Focus:

What is the focus of the following parabola equation `y^ (2) = 8x` ?

Solution:

Given parabola equation is `y^ (2) = 8x` is of the form y`^ (2) = 4ax`

We know that the formula for focus, `p = 1 / (4a)`

Now compare the given equation y2 = 8x with the general equation `y^ (2) = 4ax` . So, that `4a = 8`

From this,` p = 1 / (4a) = 1 / 8`

So, the focus of a parabola equation is `(0, 1/8)` .

Other Example Problems to Plug in and Solve Parabolas

Example 3 for Plug in and Solve Parabolas – Axis of Symmetry:

What is the axis of symmetry for parabola equation `y = 5x^ (2) + 15x + 12` `?`

Solution:

Given parabola equation is `y = 5x^ (2) + 15x + 12`

From the above equation, plug `a = 5` and `b = 15` in the axis formula.

So the axis of the symmetry of the given parabola is `-b/ (2a) = - 15/ (2 xx 5) = -15/10 = - 3/2`

Therefore, the axis of symmetry for a given parabolic curve equation is` -3/2` .

Example 4 for Plug in and Solve Parabolas – Latus Rectum:

Find the latus rectum of the given parabola equation `y^ (2) = 12x` .

Solution:

The given parabola equation is `y^ (2) = 12x`

To find the latus rectum, we have to find the value of` p` .

The parabola equation is of the form `y^ (2) = 4ax`

Here `4a = 12`

So, `p = 1/ (4a) = 1/12`

The formula for latus rectum is `4p` .

Plug the value for `p =1/12` in the latus rectum formula.

From this, the latus rectum of the parabola is `= 4p = 4 (1/12) = 4/12 = 1/3`

Therefore, the latus rectum for the parabola equation is `1/3` .

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}.

Example of Congruence

Introduction to congruence:

Two objects are congruent if they consist of the similar shape with size. The given two triangles are congruent if their equivalent sides are equal within length also their equivalent angles are the same in size. Assume the triangles DEF and RST is congruent. These can be written as ? DEF ? ? RST. In geometry two congruent triangles contains the equal corresponding angles.

Example of Congruence

The followings are the important congruence test

ASA congruence

The two angles with the integrated faces of one triangle are equal to the corresponding two angles with the integrated faces of another triangle.

SAS congruence

The two faces with the included angle of triangle are the same to two faces with the included angle of another triangle.

AAS congruence

The two angles along throughout a non integrated side of one triangle are congruent to the identical measurements of a different triangle.

SSS congruence

Three sides of one triangle are identical to corresponding three sides of another triangle.

Example

Consider the following two triangles.




The triangle IJK is congruent to the triangle LMN

Angle I = Angle L

Angle J = Angle M

Angle K = Angle N

Length IJ = Length LM

Length JK = Length MN

Length KI = Length LN

Understanding how do you simplify fractions is always challenging for me but thanks to all math help websites to help me out.

Examples for Congruence

Example 1 for congruence

In the following triangles are congruent then find the length of sides a, b, c.



Solution

The given triangles are congruent. Therefore the lengths of the sides of the triangles are equal.

Length EG = 52

Therefore the length VT = a = 52

Length FG = 48

Therefore the length UT = b = 48

Length EF = 50

Therefore the length UV = c = 50

Thus the a = 52, b = 48, c = 50.

Example 2 for congruence

Prove that triangle LMN is congruent to triangle PQR.



Solution

Given figure the angle L and angle P are the same.

Angle L = Angle P = 75 degree

Given figure the angle N and angle Q are the same.

Angle N = Angle Q = 65 degree.

Line segment LM is equal to the line segment PR.

Line LM = Line PR = 40 cm.

Therefore ? LMN and ? PQR are congruent through AAS congruence.

Unit Circle Equation

Introduction :

A unit circle is defined as a circle with the radius value is one. Particularly in the trigonometry the unit circle with radius one is pointed at (0, 0) that is the origin in Euclidean plane of the Cartesian coordinate system. The unit circle is represented as S1. The higher dimension of the unit circle is called as the unit sphere.

Formula for Unit Circle Equation:

If the point (x, y) is on the first quadrant of the unit circle, then the point x and y are the lengths of the right triangle and the hypotenuse length is one. By using the Pythagorean Theorem, the equation of the unit circle is,

` x^2 +y^2=1`

consider` x^2=(-x)^2` for all the value of x ,it gives the positive x value and the reflection of any point of x and y axis of the unit circle is provides the unit circle equation that is ` x^2 +y^2=1` and this is not only for the first quadrant of the unit circle.

The unit circle coordinates:

The unit circle having the angle theta and also having the radius one for the unit circle. The unit circle coordinates are,(x,y) that is,

`x=cos theta or cos theta=x/1=x`

`y=sin theta or sin theta=y/1=y`

By using the Pythagorean Theorem, the equation of the unit circle is,

`cos^2 theta + sin^2 theta=1`

Example 1 for Unit Circle Equation:

To check whether the following points are on the unit circle equation or not.

i) ` (1/ 2, sqrt3/2)`

Solution:

Take the unit circle equation is,

`cos^2 theta + sin^2 theta=1`

`x^2 +y^2=1`

` x=1/2 and y=sqrt 3/2`

put x and y values in the unit circle equation

`(1/2)^2+(sqrt 3/2)^2 =1/4 +3/4 =1`

Therefore these two points are situated on the unit circle equations.

`cos theta =1/2`

`theta =cos^(-1) (1/2) =pi/3 =60^@`

Therefore these two points are situated on the unit circle equations with the angle `theta=60^@.`

Example 2 for unit circle equation:

To check whether the following points are on the unit circle equation or not.

i) ` (0, 1)`

Solution:

Take the unit circle equation is,

`cos^2 theta + sin^2 theta=1`

`x^2 +y^2=1`

here x=0 and y=1

put x and y values in the unit circle equation

`(0)^2+(1)^2 =0 +1 =1`

Therefore these two points are situated on the unit circle equations.

`cos theta =0`

`theta =cos^(-1) (0) =pi/2 =90^@`

Therefore these two points are situated on the unit circle equations with the angle `theta=90^@.`

Three Dimensional Pyramid

Introduction:

In geometry, we have two-dimensional shape and three dimensional shape. Pyramid is a thee dimensional shape. It is also called as Polyhedron. The base of pyramid may be square, rectangle, triangle or any other polygon shape. The sides of pyramid look triangular faces. In Pyramid, all vertices of base are connecting a point to above, it is called apex of pyramid.

We have various kind of three dimensional pyramids. They are following.

For example, we have Square based pyramid, Rectangular pyramid, Triangular Pyramid, Pentagonal pyramid, Hexagonal pyramid and etc.

In this article, we will see about volume of all kinds of three dimensional pyramid.

Three Dimensional Pyramid: Square Pyramid, Rectangle Pyramid

Square based pyramid:

It is a three dimensional pyramid shape having a base is square. In square pyramid, we have five vertices, four triangular faces.

Formula:

Volume of square pyramid = `1/3*base^2*height`

Example 1:

Find the volume of square pyramid for the given base side is 7 meter and height is 10 meter.

Solution:

Given:

Base side = 7 m

Height = 10 m

Volume of square pyramid           = `1/3*base^2*height`

= `1/3*7^2*10`

= 0.333 * 49 * 10

= 163.17

Therefore, Volume of square pyramid is 163.2 cubic meter.

Rectangular pyramid:

It is a three dimensional pyramid shape having a base is Rectangle. In Rectangle pyramid, we have five vertices, four triangular faces.

Formula:

Volume of Rectangular pyramid = `1/3*LenGth*width*height`

Example 2:

Find the volume of Rectangular pyramid for the given length is 10 meter , width is 8 meter and height is 12 meter.

Solution:

Given:

Length = 10 m

Width = 8 m

Height = 12 m

Volume of rectangular pyramid          = `1/3*LenGth*width*height`

= `1/3*10*8*12`

= 0.333 * 960

= 319.68

Therefore, Volume of Rectangular pyramid is 319.7 cubic meter.

Three Dimensional Pyramid: Triangular Pyramid, Hexagonal Pyramid

Triangular pyramid:

It is a three dimensional pyramid shape having a base is Triangle. In triangular pyramid, we have four vertices, four triangle faces.

Formula:

Volume of triangular pyramid = `1/6*base*height*HEIGHT`

Example 3:

Find the volume of triangular pyramid for the given base is 6 meter , height is 10 meter and HEIGHT is 12 meter.

Solution:

Given:

Base = 6 m

Height = 10 m

HEIGHT = 12 m

Volume of Triangular pyramid      = `1/6*base*height*HEIGHT`

= `1/6*6*10*12`

= 0.167 * 720

= 120.24

Therefore, Volume of triangular pyramid is 120.2 cubic meter.

Hexagonal pyramid:

It is a three dimensional pyramid shape having a base is Hexagon. In Hexagonal pyramid, we have seven vertices, six triangle faces.

Formula:

Volume of Hexagonal pyramid = `Apothem*Side*Height`

Example 3:

Find the volume of Hexagonal pyramid for the given side is 6 meter , height is 10 meter and Apothem is 7 meter.

Solution:

Given:

Side = 6 m

Height = 10 m

Apothem = 7 m

Volume of Hexagonal pyramid     = `Apothem*Side*Height`

= `7*6*10`

= 420

Therefore, Volume of Hexagonal pyramid is 420 cubic meter.

Solving Geometry Explanation

Introduction :-

In geometry, an arc is a segment of a differentiable curve in the two-dimensional plane; for example, a circular arc is a segment of the circumference of a circle. If the arc segment occupies a great circle (or great ellipse), it is considered a great-arc segment.I like to share this Math Pythagorean Theorem with you all through my article.
(Source : Wikipedia)

Example Problems for Solving Geometry Explanation

Problem 1:-

Solving geometry explanation to find the volume of cone with radius 7 cm and height 8 cm.

Solution:

Given: Radius = 7 cm

Height = 8 cm.

Volume of cone = (`1/3` ) * `pi` * radius2 * height

= (`1/3` ) * 3.14 * 72 * 8  ( multiplying these values)

= 0.33 * 3.14 * 49 *9  ( multiplying the values)

= 456.96 cubic cm.

The volume of cone is 456.96 cubic cm.

Problem 2:

Solving geometry explanation to find the Perimeter of Parallelogram for the side of a 8 and side of b is 6.

Solution:

Given: Side a = 8

Side b = 6

Perimeter of Parallelogram P = 2 * 8 + 2 * 6  ( multiplying the values)

P = 16 + 12

P = 28

The Perimeter of Parallelogram is 32

Problem 3:

Solving geometry explanation to find the circle area and circumference radius with 6 cm.
Solution:

Given: Radius = 6 cm

Area of Circle = `pi` * radius2          `pi` = 3.14

= 3.14 * 62

= 3.14 * 36   ( multiplying the values)

= 113.04 square cm.

The Area of Circle is 113.04 square cm

Circumference of Circle = 2 * `pi` * radius

= 2 * 3.14 * 6   ( multiplying the values)

= 37.68cm.

The Circumference of Circle 37.68 cm

More Example Problems for Solving Geometry Explanation

Problem 1:

Solving geometry explanation to find the Area of Triangle with height 3 cm and Base 7 cm.

Solution:

Given: Height = 3 cm

Base = 7 cm

Area of Triangle = (½) * height * base

= 0.5 * 3* 7   ( multiplying these values)

= 10.5 square cm.

The Area of Triangle 10.5 square cm

Problem 2:

Solving geometry explanation to find the Area of rhombus whose diagonal lengths are 5 cm and 8 cm.

Solution:

Area of Rhombus = (½) * Length of the diagonal 1 * Length of the diagonal 2

= `1/2` * 5* 8 ( multiplying these values)

= 20 square cm.

The Area of Triangle 20 square cm

Line Segments in a Pentagon

Introduction to line segments:

The division of a line with two end points is called a line segment. Line segment RS which we denoted by the symbol `bar(RS)` .



Note: We shall denote a line segment `bar(RS)` by RS only.

From the above figure, we call it a line segment RS. The points R and S are called end-points of the line segment RS.

We can also name it as line segment RS.

A line segments:

(a) A line segment has a definite length.

(b) A line segment has two end-points

Line Segments in a Pentagon:
Find the line segments of the given pentagon. The pentagon shown below figure,



Solution:

Given:

Pentagon EFGHI

To find the line segments in a pentagon:

We know that the line segments are consisting of two end points. Here, the pentagon has five end points, such as E, F, G, H, and I. The five end points to form the line segments in a pentagon by connecting these end points shown in figure, such line segments are EF, FG, GH, HI, and IE. These line segments are represented by `bar(EF)` , `bar(FG)`, `bar(GH)` , `bar(HI)` , and `bar(IE)` . Therefore, the given pentagon has five line segments.Please express your views of this topic Converting Fractions to Percents by commenting on blog.

Line Segments in a Solid Pentagon:

Find the line segments of the given solid pentagon. The solid pentagon shown below figure,



Solution:

Given:

Solid pentagon ABCDEFGHIJ

To find the line segments in a solid pentagon:

We know that the line segments are consisting of two end points. Here, the pentagon has ten end points, such as A, B, C, D, E, F, G, H, I, and J. The ten end points to form the line segments in a solid pentagon by connecting these end points shown in figure, such line segments are AB, AD, AJ, BC, BF, CD, CE, DI, EF, EG, FH, GH, GI, HJ, and IJ. These line segments are represented by `bar(AB)` , `bar(AD)` , `bar(AJ)` , `bar(BC)` ,` bar(BF)` , `bar(CD)` , `bar(CE)` , `bar(DI)` , `bar(EF)` , `bar(EG)` , `bar(FH)` , `bar(GH)` , `bar(GI)` , `bar(HJ)` , and `bar(IJ)` . Therefore, the given solid pentagon has fifteen line segments.

Radius of a Circle from Circumference

Radius of a circle from circumference:
The terms radius, diameter and circumference are related to two-dimensional geometric shape named circle. Circle is a two dimensional closed shape with curved edges. The distance between the center of the circle and any point on  the circle  is always same. Circumference of the circle is 2pi r  .where, r is the radius of the circle .

Here radius is the distance between the center of the circle to any point on the circles. Circumference is the total distance around the circle. Let us discuss about the radius from the circumference of the circle,



Example Problem to Find Radius from Circumference:

Example 1:

Find radius of the circle if circumference is 34cm.

Solution:

The classic formula for circumference is `2 pi r`

Therefore,

Circumference =2`pi` r =34cm

Simplify it for radius (r) we get ,

Radius r=`34/2pi`

We know that `pi ` =3.14 or` 22/7`

Therefore, r= `34/(2*3.14)`

=5.41cm

Therefore value of radius from circumference is 5.41cm

Example 2:

Find radius of the circle if circumference is 23cm.

Solution:

The classic formula for circumference is `2 pi r`

Therefore,

Circumference =`2pir` =23cm

Simplify it for radius (r) we get ,

Radius `r=23/2pi`

We know that pi =3.14 or `22/7`

Therefore,` r= 23/(2*3.14)`

=3.66cm

Therefore, value of radius from circumference is 3.66 cm

Example 3:

Find radius of the circle if circumference is 72.4cm.

Solution:

The classic formula for circumference is` 2 pi r`

Therefore,

Circumference =`2pir ` =72.4cm

Simplify it for radius (r) we get ,

Radius r=`72.4/(2pi)`

We know that `pi` =3.14 or `22/7`

Therefore, r=`72.4/(2*3.14)`

=11.52 cm

Therefore, value of radius from circumference is 11.52cm

Example 4:

Find radius of the circle if circumference is 11cm.

Solution:

The classic formula for circumference is `2 pi r`

Therefore,

Circumference =`2pir` =11cm

Simplify it for radius (r) we get ,

Radius r=`11/2pi`

We know that pi =3.14 or `22/7`

Therefore, r= `11/(2*3.14)`

=1.75cm

Therefore, value of radius from circumference is 1.75 cm

Example 5:

Find radius of the circle if circumference is 28inch.

Solution:

The classic formula for circumference is `2 pi r`

Therefore,

Circumference =`2pir ` =28inch

Simplify it for radius (r) we get ,

Radius` r=28/(2pi)`

We know that `pi` =3.14 or `22/7`

Therefore, r= `28/(2*3.14)`

=4.45inch

Therefore, value of radius from circumference is 4.45inch

Is this topic how to measure volume hard for you? Watch out for my coming posts.

Practice Problem to Find Radius from Circumference:

1) Find radius of the circle if circumference is 12cm.

Answer:1.91cm

2)Find radius of the circle if circumference is 44cm.

Answer:7cm

Trisecting a Line Segment

Introduction for line segment:

A line segment is the basic and fundamental topic in geometry and math subject. Generally in math, a straight and long line is divided with two definite end points on both sides are known as line segments. Here in this article we have to brief explain about line segment and trisecting a line segment. And use dome example figures for how to do trisecting line segment.

Line Segment General Definition:

In math, a line segment is can be defined as one small part or distance between the two endpoints of a long line.
Line segment is also shape like as ‘a straight line’, which is joining the two points with coordinates, and the line is infinity after that, the end points.
Example figure for general line segment:



In this figure xy is the infinity line, and A, B are the two end points, and the line segments are` bar (AB)` .
The length of the line segments AB would be written as `bar (AB)` . And the line segments have used the name as to be two end points AB.

Trisecting a Line Segment:

Trisecting a line segment is the one of the process of dividing the line segments with a new line and makes the new line segments.  It is simply defined as which is one line segment, is trisected.

First we have to draw a line segment, and then bisect the line segment with a new line, and then we have to get trisecting line segment.

Generally trisecting a line segment, we use compass and ruler and makes easy.

Step by step process for trisecting line segment:

Step 1:

First we have to draw a line segment with two end points A, B. And name of the line segment is AB.

Stwp2:

Then next, draw a new dotted line through endpoint A, but not coincident with AB, draw that line for our convenient and put an end point with the name of  C and D.

Step 3:

Here the line segment distances of AC and AD are equal.

Step 4:

Then again draw a dotted line through End point B, same length and put end points and named as E and F.

This E and F are opposite lines for AB from points C and D. distance BE = EF.

Step 5:

Connect CF and DE; those two lines will cut AB into third lines equal.

Example figure for trisecting line segment:



The above example figure and explanations will make clears for the trisecting a line segment.

Solving Centroid Formula

Introduction to solve centroid formula:

In general, Centroid formula is a point on a given body or shape at which the entire mass of the body acts (center of gravity of the mass), it might also be the center of area for certain shapes. For a triangle, solving centroid is the point at which the medians of the triangle intersect; they intersect at the ratio 2:1. In the case of polygons the Centroid is found using the boundary co-ordinates solving.

Solving Centroid Formulae:

In this case the Centroid of the triangle is taken and the formula used to find out solving the centroid of a triangle is,

G (x1+x2+x3)/3 , (y1+y2+y3)/3
Where,
(x1, y1)
(x2, y2)
(x3, y3)
are the co-ordinates of the triangle.

In general, for any shape in the x-y plane the Centroid formulae can be generalized to,
G (x1+x2+x3+….+xn)/3n , (y1+y2+y3+….+yn))/n
Where,
(x1, y1)
(x2, y2)
(x3, y3)
(........)
(........)
(xn, yn) are the co-ordinates of the given shape.

Example Problems on Solving Centroid Formula:

1. Calculate the Centroid of a triangle whose co-ordinates are (3, 6) (4, 2) (3, -4)

Sol:
The given points are (3, 6) (4, 2) (3, -4),
Therefore solving,
(x1, y1) is (3, 6)
(x2, y2) is (4, 2)
(x3, y3) is (3, -4)

Formulae for the Centroid of triangle is,
G (x1+x2+x3)/3 , (y1+y2+y3)/3
(3+4+3)/3 , (6+2-4)/3
(10)/3 , (8-4)/3
10/3 , 4/3
3.33 , 1.33
Therefore the Centroid is (3.33, 1.33)

2. Calculate the Centroid of a triangle whose co-ordinates are (4, 8) (3, 2) (5, -4)
Sol:
The given points are (4, 8) (3, 2) (5, -4),
Therefore solving,
(x1, y1) is (4, 8)
(x2, y2) is (3, 2)
(x3, y3) is (5, -4)

Formulae for the Centroid of triangle is,
G (x1+x2+x3)/3 , (y1+y2+y3)/3
(4+3+5)/3 , (8+2-4)/3
(12)/3 , (10-4)/3
12/3 , 6/3
4 , 2
Therefore the Centroid is (4, 2).

3. Calculate the Centroid of the quadrilateral, whose co- ordinates are (3, 2) (5, -4) (4, 2) (3, -4)

Sol:
The given points are (3, 2) (5, -4) (4, 2) (3, -4),
Therefore,
(x1, y1) is (3, 2)
(x2, y2) is (5, -4)
(x3, y3) is (4, 2)
(x4, y4) is (3, -4)

Formulae for the Centroid is
G (x1+x2+x3+….+xn)/3n , (y1+y2+y3+….+yn))/n,
(3+5+4+3)/4 , (2-4+2-4)/4,
(15)/4 , (4-8)/4,
3.75 , -4/4
3.75 , -1
Therefore the Centroid is (3.75, -1)

Three Horizontal Lines

Introduction to three horizontal lines:

Three horizontal lines are nothing but three lines parallel to x – axis or three lines perpendicular to y – axis. Three horizontal lines mean their slopes will be zero. Because the slope of x – axis is 0. We know if there is any two lines are parallel their slope s will be equal. We will some example problems for graphing three horizontal lines. If the line is horizontal their y value is constant.

Examples for three Horizontal Lines:
If the line is parallel to x – axis we can say those lines are horizontal lines. The slopes of the horizontal lines are zero and the y value of the line is constant. So the equation of the horizontal lines are like y = some constant value.Having problem with geometric probability formula keep reading my upcoming posts, i will try to help you.

Example 1 for three horizontal lines:

Graph the following lines y = 1, y = 5 and y = -1.

Solution:

Here the line equations are y = 1, y = 5 and y = -1

The slope intercept form general equation is y = mx + c

If we compare the given equation with general form we can get the slope of the lines are 0.

If we graph these equations we will get the graph like the following.



More Examples for three Horizontal Lines:
Example 2 for three horizontal lines:

Graph the following lines y = 2, y = 3 and y = -2.

Solution:

Here the line equations are y = 0, y = 3 and y = -2

The slope intercept form general equation is y = mx + c

If we compare the given equation with general form we can get the slope of the lines are 0.

If we graph these equations we will get the graph like the following.



In this the line y = 0 is lies on the x axis.

These are some of the examples for three horizontal lines. From the above we can understand how to graph the three horizontal lines and slopes of the horizontal lines.