A student wants to find point C on the directed line segment from A to B on a number line such that the segment is partitioned in a ratio of 3:4. Point A is at
Question
2 Answer

1. User Answers Matheng
Answer:
Point C at (18/7)
Stepbystep explanation:
Point A is at 6 and point B is at 2.
So, the distance between A and B = 2  (6) = 8
point C on the directed line segment from A to B on a number line such that the segment is partitioned in a ratio of 3:4.
Let the distance AC = x
∴ BC = 8  x
AC : CB = 3 : 4
∴[tex]\frac{AC}{CB} =\frac{3}{4} = \frac{x}{8x}[/tex]
Using cross multiplication
3 (8x) = 4x
24  3x = 4x
24 = 7x
x = 24/7
So, Point C = 6 + 24/7 = 18/7
Point C at (18/7)

2. User Answers psm22415
The point C directed line segment from A to B is 18/7.
Given that,
A student wants to find point C on the directed line segment from A to B on a number line such that the segment is partitioned in a ratio of 3:4.
Point A is at 6 and point B is at 2.
We have to determine,
The point C on the directed line segment from A to B.
According to the question,
Let, the distance AC be x,
And BC = 8  x
The ratio of AC : BC = 3 : 4
Therefore,
[tex]\dfrac{AC}{BC} = \dfrac{3}{4} = \dfrac{x}{8x}[/tex]
Solving the equation by cross multiplication,
[tex]3 (8x) = 4x\\\\24  3x = 4x\\\\24 = 7x\\\\x = \dfrac{24}{7}[/tex]
Then,
Point C directed line segment from A to B is,
[tex]= 6 + \dfrac{24}{7}\\\\= \dfrac{42+24}{7}\\\\= \dfrac{18}{7}[/tex]
Hence, The point C directed line segment from A to B is 18/7.
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