Length of Lines

Introduction:

The Length of lines is the main topic in Geometry. The line length between two points, it is used in all geometric concepts. Let P1 (x1, y1) and P2 (x2, y2) be two distinct points in the Cartesian plane and denote the line length between P1 and P2 by d (P1, P2) or by P1P2. The length formula is used to find the length between two points.  With the help of geometric properties and the length formula we have to identify the given points make which shape.


Brief Description of Line Length Formula

Consider the line segment P1 P2. Here two cases arise.

Case (I):

The segment P1 P2 is parallel to the x-axis Then y1 = y2. Draw P1L and P2M, perpendicular to the x-axis. Then d (P1, P2) is equal to the length between A and B. But A is (x1, 0) and B is (x2, 0). So the length AB =|x1 -X2|. Then d (P1, P2) =|x1 -X2|.

Case (II):

The segment P1 P2 is parallel to the y-axis. Then x1 = x2 .Draw P1A and P2B, perpendicular to the y-axis. Then d (P1, P2) is equal to the line length between A and B. But A is (0, y1) and B is (0, y2). So the length AB =| y1- y2|.then d (P1, P2) = | y1- y2|

`sqrt(((x2-x1)^2 + (y2-y1)^2))`

This is called the line length formula which gives the line length d between the two given points (x1, y1) and (x2, y2). We observe that d (P1, P2) = d (P2, P1). The formula has been derived for two points which are not on a horizontal line or vertical line.

Examples Problems

Given points are P (5,4) and Q (8, 6). Find the length of PQ using line length formula.

Solution:

Let d be the line length between P and Q. Here we have to find the length between two points using line length formula.

Then d (P, Q) =`sqrt(((x2-x1)^2 + (y2-y1)^2))`

= `sqrt(((8-5)^2 + (6-4)^2))`

Simplifying this we get the length is v13.

Example 2:

Given points are W(2, -3), X(6, 5), Y(-2, 1) and Z(-6, -7), Show that this points make a square.

Solution: One way of showing that WXYZ is a rhombus is to show that all its sides are of equal length. One way is showing that a rhombus is not a square is to show that the diagonals are of unequal length. Here we find

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

Simplifying this we get v80

XY = `sqrt(((-2-6)^2 + (1-5)^2))`

Simplifying this we get v80

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

Simplifying this we get v32

XZ = `sqrt(((-6-6)^2 + (-7-5)^2))`

Simplifying this we get v288

YZ = `sqrt(((-6+2)^2 + (-7-1)^2))`

Simplifying this we get v80

WZ= `sqrt(((-6-2)^2 + (-7+3)^2))`

Simplifying this we get v80

Therefore WX=XY=YZ=ZW

WY is not equal to XZ

Therefore WXYZ is a rhombus but not a square.

My forthcoming post is on solving linear equations by graphing, a list of prime numbers will give you more understanding about Algebra.