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.