HOW TO FIND THE LENGTH OF A LINE SEGMENT WITH COORDINATES
To find the distance between two points, we will use the formula given below.
Distance between two points = length of the line segment
The endpoints can be considered as the following.
Calculate the length of the line joining the points A and B.
Problem 1 :
Solution:
Given, A(1, 1) and B(4, 5)
distance=(x2-x1)2+(y2-y1)2Here x1=1,y1=1,x2=4 and y2=5d=(4-1)2+(5-1)2=32+42=9+16=25d=5
Problem 2 :
Solution:
Given, A(1, 0) and B(3, 6)
distance=(x2-x1)2+(y2-y1)2Here x1=1,y1=0,x2=3 and y2=6d=(3-1)2+(6-0)2=22+62=4+36=40d=210
Problem 3 :
Solution:
Given, A(1, 4) and B(6, 3)
distance=(x2-x1)2+(y2-y1)2Here x1=1,y1=4,x2=6 and y2=3d=(6-1)2+(3-4)2=52+(-1)2=25+1=26d=213
Calculate the length of the line joining the point A and B.
Problem 4 :
Solution:
Given, A(7, 12) and B(12, 0)
distance=(x2-x1)2+(y2-y1)2Here x1=7,y1=12,x2=12 and y2=0d=(12-7)2+(0-12)2=52+(-12)2=25+144=169d=13
Problem 5 :
Solution:
Given, A(2, 3.6) and B(6.5, 5.2)
distance=(x2-x1)2+(y2-y1)2Here x1=2,y1=3.6,x2=6.5 and y2=5.2d=(6.5-2)2+(5.2-3.6)2=(4.5)2+(1.6)2=20.25+2.56=22.81d=4.77
Problem 6 :
Calculate the perimeter of triangle ABC.
Solution:
distance AB=(x2-x1)2+(y2-y1)2Here x1=-3,y1=-1,x2=-1 and y2=5AB=(-1+3)2+(5+1)2=(2)2+(6)2=4+36=40AB=210BC=(4+1)2+(-2-5)2=(5)2+(-7)2=25+49BC=74AC=(4+3)2+(-2+1)2=(7)2+(-1)2=49+1=50AC=52
Perimeter = AB + BC + AC
= 2√10 + √74 + 5√2
Problem 7 :
The distance between the points (1, 2) and (16, p) is 17. Find the possible values of p.
Solution:
distance =(x2-x1)2+(y2-y1)2Here x1=1,y1=2,x2=16 and y2=p17=(16-1)2+(p-2)2Squaring on both sides,172=152+(p-2)2289=225+p2+4-4pp2+4-4p+225-289=0p2-4p-60=0(p+6)(p-10)=0p=-6 or p=10
Problem 8 :
The distance between the points (-3, -4) and (q, 5) is 15. Find the possible values of q.
Solution:
distance =(x2-x1)2+(y2-y1)2Here x1=-3,y1=-4,x2=q and y2=515=(q+3)2+(5+4)2Squaring on both sides,152=(q+3)2+92225=q2+9+6q+81q2+9+6q+81-225=0q2+6q-135=0(q+15)(q-9)=0q=-15 or q=9