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Example of finding the length of a part of a line segment
Question:
In the figure below. the ratio of the segment AB to BC is 2 : 3. If AC is 150 feet, what is the distance from the point A to the point that is 2/3 of the distance from B to C? Solution:
since AB : BC = 2 : 3 (Given)
that is, AB/BC = 2/3
so AB = (2/3)BC
let BC = 3x and AB = 2x
note: AB/BC = 2x/3x = 2/3
since points A, B, and C are all on the same line
so AB + BC = AC
Given AC = 150
so AB + BC = 150
express AB and BC in terms of x
2x + 3x = 150
5x = 150
x = 150/5 = 30
so AB = 2x = 2(30) = 60
so BC = 3x = 3(30) = 90
let D is the point that is 2/3 of the distance from B to C
BD = (2/3)BC
then BD = (2/3)90 = 60
AD = AB + BD = 60 + 60 = 120
therefore, the distance from A to the point that is 2/3 of the distance from B to C is 120 feet.

This is a ratio problem. Given the ratio of AB to BC is 2 to 3. That is, AB divide by BC is equal to = 2 divide by 3. So AB is equal to two third of BC. Let BC is three times the value of x, then AB is equal to two times x. From the figure given, the length of AB plus the length of BC is 150. So we get the equation 2x + 3x = 150. Combine Similar (like) terms to make this equation into a standard form ax = b, then x = b/a. Now we get the value of x. BC = 3x and AB = 2x. Now we get the values of AB and BC. Let D is the point that is 2/3 of the distance from B to C. Then BD = (2/3)BC. AD = AB + BD. Thus we get the distance AD.