Question

Explain what a direct line segment is and describe how you would find the coordinates of point P along a directed line segment AB that partitions segment AB so that the ratio of AP to PB is 3:1

Answer 1

Answer:

  P = (A +3B)/4

Step-by-step explanation:

A "directed" line segment is one that has a "beginning" and an "end". The first letter of the line segment's name is the beginning; the last letter is the end.

Line segment AB begins at point A and ends at point B.

The problem statement tells you what it means to locate point P so the segment is divided in the ratio 3:1.

  AP : PB = 3 : 1

  (P -A) / (B -P) = 3 / 1

  P -A = 3(B -P) . . . . . . . . multiply by (B-P)

  P +3P = A +3B . . . . . . . add A+3P

  P = (A +3B)/4

You can find point P by filling in the coordinate values in the formula just found.

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Comment on the general case

If you derive the formula for P dividing the segment in the ratio m : n, you find it is ...

  P = (nA +mB)/(m+n) . . . length ratios are applied in reverse to the end points