MATH SOLVE

3 months ago

Q:
# Let AB be the directed line segment beginning at point A(3 , 2) and ending at point B(-12 , 10). Find the point P on the line segment that partitions the line segment into the segments AP and PB at a ratio of 4:5

Accepted Solution

A:

Answer:The coordinates of point P are: Β [tex]\displaystyle \left(-\frac{11}{3}, \frac{50}{9}\right)[/tex].Step-by-step explanation:[tex]\rm |AB| = |AP| + |PB|[/tex].Therefore, if [tex]\rm |AP|:|PB| = 4 :5[/tex], then[tex]\rm |PB| : |AB| = |PB| : (|AP| + |PB|) = 5: (4 + 5) = 5 : 9[/tex].Consider: what's the horizontal and vertical separation between point A and point B?Horizontal SeparationNote that the [tex]x[/tex]-coordinate (the first of the two) of point B is smaller than that of point A by [tex]3 - (-12) = 15[/tex]. In other words, point A is to the right of point B with a horizontal separation of [tex]15[/tex] units. However, since point P is somewhere between point A and B, it should also also be to the left of point B. Additionally, since [tex]\rm |PB|: |AB| = 5 : 9[/tex], point P should be to the left of point B with a horizontal separation of [tex]\displaystyle 15 \times \frac{5}{9} = \frac{25}{3}[/tex] units.As a result, the horizontal coordinate of point P would be [tex]\displaystyle \displaystyle -12 + \frac{25}{3} = - \frac{11}{3}[/tex].Vertical SeparationSince the [tex]y[/tex]-coordinate (the second of the two) of point B is larger than that of point A by [tex]10 - 2[/tex] units, point B is above point A with a vertical separation of [tex]8[/tex] units.Since point P is between point A and B, it should also be above point A and below point B. P should be below point B with a vertical separation of [tex]\displaystyle 8 \times \frac{5}{9} = \frac{40}{9}[/tex].As a result, the vertical coordinate of point P would be equal to [tex]\displaystyle 10 - \frac{40}{9} = \frac{50}{9}[/tex].Overall, the coordinates of point P should be [tex]\displaystyle \left(-\frac{11}{3}, \frac{50}{9}\right)[/tex].