Points and lines tutoring

Tutor is the person who teaches the kids and this teaching section is the tutoring. Tutoring is an open source for the students to gain knowledge that is in online a point is nothing but the dot , it has no dimension or no width, it’s only a simple black dot. In geometry co ordinates of a point which shows the particular place in a segment for representation.Line has two end points is called segment. Line segment is denoted with a connected piece of line.line segments names  has two endpoints and it is named by its endpoints.

Points and Lines Tutoring:

Tutoring about the geometric points and lines we have to know the classification of a points and lines. points and lines classification are as follows.

Collinear points:
When three or more points lies on the same line is said to be collinear points.

Midpoint:
A halfway point where line segment divides into two equal parts are called midpoint.

Equidistant point:
A point which is said to be equidistant in a line segment where point is equal length from other points which are in congruent then the point is equidistant point.

Parallel line segment:
Two lines which does not touch each other are called parallel lines.

Perpendicular line segment:
Two line segment  that form a L shape are called perpendicular lines.

Problems in Points and Lines Tutoring:

Example 1:
Find the distance between the points A(6,3) and B (2,1).

Solution:
Let assume "d" be the distance between A and B.           (x1,y1)= (6,3), (x2,y2)= (2,1).

Then d (A, B) =`sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

= `sqrt((2-6)^2 +(1-3))^2)`

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

= `sqrt(16+4)`

=`sqrt20`
=2`sqrt5`

Example 2:
Find co-ordinate of the mid point of the line segment joining given points A(-5,3) and B(2,1)

Solution:
The required mid point is
Formul a   `((x_1+x_2)/2 ,(y_1+y_2)/2)` here,  (x1, y1) = (-5,3),(x2, y2) = (2,1)

=  `((-5+2)/(2))``((3+1)/(2)) `

=   `(-3/2) ` , ` (4/2)`

=    `(-3/2, 2)`

Example 3:
Find the slope of the lines given (8,-5) and (4,2)

Solution:
(x1,y1)= (8,-5), (x2,y2)= (4,2).
We know to find slope of line,m=` (y_2-y_1) /(x_2-x_1)`

=`(2+5)/(4-8)`

m =`7/-4`

Example 4:
Find the equation of the line having slope  3 and y-intercept 5.

Solution:
Applying the equation of the line is y = mx + c
Given,       m =3 ,c = 5
y =  3x +5

or  y = 3x+5
or  -3x+y-5 = 0
3x-y+5 = 0.