Respuesta :
We have a one-dimensional horizontal line segment. Three points are indicated on the line as follows:
In the above sketch we have first denoted a reference point at the extreme left hand as ( Ref = 0 ). This is classified as the origin. The point ( A ) is located on the same line and is at a distance of ( 21 units ) from Reference ( Ref ). The point ( B ) is located on the same line and is at a distance of ( 66 units ) from Reference ( Ref ).
The point is located on the line segment ( AB ) in such a way that it given as ratio of length of line segment ( AB ). The ratio of point ( P ) from point ( A ) and from ( P ) to ( B ) is given as:
[tex]\textcolor{#FF7968}{\frac{AP}{PB}}\text{\textcolor{#FF7968}{ = }}\textcolor{#FF7968}{\frac{3}{2}\ldots}\text{\textcolor{#FF7968}{ Eq1}}[/tex]The length of line segment ( AB ) can be calculated as follows:
[tex]\begin{gathered} AB\text{ = OB - OA } \\ AB\text{ = ( 66 ) - ( 21 ) } \\ \textcolor{#FF7968}{AB}\text{\textcolor{#FF7968}{ = 45 units}} \end{gathered}[/tex]We can form a relation for the line segment ( AB ) in terms of segments related to point ( P ) as follows:
[tex]\begin{gathered} \textcolor{#FF7968}{AB}\text{\textcolor{#FF7968}{ = AP + PB }}\textcolor{#FF7968}{\ldots}\text{\textcolor{#FF7968}{ Eq2}} \\ \end{gathered}[/tex]We were given a ratio of line segments as ( Eq1 ) and we developed an equation relating the entire line segment ( AB ) in terms two smaller line segments as ( Eq2 ).
We have two equation that we can solve simultaneously:
[tex]\begin{gathered} \textcolor{#FF7968}{\frac{AP}{PB}}\text{\textcolor{#FF7968}{ = }}\textcolor{#FF7968}{\frac{3}{2\text{ }}\ldots}\text{\textcolor{#FF7968}{ Eq1}} \\ \textcolor{#FF7968}{AB}\text{\textcolor{#FF7968}{ = AP + PB }}\textcolor{#FF7968}{\ldots Eq2} \end{gathered}[/tex]Step 1: Use Eq1 and express AP in terms of PB.
[tex]AP\text{ = }\frac{3}{2}\cdot PB[/tex]Step 2: Substitute ( AP ) in terms of ( PB ) into Eq2
[tex]AB\text{ = }\frac{3}{2}\cdot PB\text{ + PB}[/tex]We already determined the length of the line segment ( AB ). Substitute the value in the above expression and solve for ( PB ).
Step 3: Solve for PB
[tex]\begin{gathered} 45\text{ = }\frac{5}{2}\cdot PB \\ \textcolor{#FF7968}{PB}\text{\textcolor{#FF7968}{ = 18 units}} \end{gathered}[/tex]Step 4: Solve for AP
[tex]\begin{gathered} AP\text{ = }\frac{3}{2}\cdot\text{ ( 18 )} \\ \textcolor{#FF7968}{AP}\text{\textcolor{#FF7968}{ = 27 units}} \end{gathered}[/tex]Step 5: Locate the point ( P )
All the points on the line segment are located with respect to the Reference of origin ( Ref = 0 ). We will also express the position of point ( P ).
Taking a look at point ( P ) in the diagram given initially we can augment two line segments ( OA and AP ) as follows:
[tex]\begin{gathered} OP\text{ = OA + AP} \\ OP\text{ = 21 + 27} \\ \textcolor{#FF7968}{OP}\text{\textcolor{#FF7968}{ = 48 units}} \end{gathered}[/tex]The point ( P ) is located at.
Answer:
[tex]\textcolor{#FF7968}{48}\text{\textcolor{#FF7968}{ }}[/tex]
