determine the point p that partitions segment ab, a(-3, -14) b (21, 2) into a 3:5 ratio

Question

thanks

Answer 1

Segment ab from point a(-3,14) to point b(21,2) has a length (right now, we don't know and don't care what it is).  We want to find point p (x.y) on that segment that divides the segment into two segments with a ratio of 3:5.

 

O.K., the segment ab has a total of 8 (3+5) sub-parts.  There are 3 in segment ap and 5 in segment pb.

 

Now, segment ab has associated with it y-differences and x-differences that may be computed --   there are two possible points that make a right triangle with points a and b such that segment ab is the hypotenuse.  They have coordinates (-3,2) and (21,14).

 

Not surprisingly, the segments connecting one of these points to a or to b will also be divided into a ratio of 3:5, so let's save the calculations of square and square root required by the Pythagorean Theorem and just find x and y.

 

From point a to point b, x goes from -3 to 21.  That's 24 units.  The values 9:15 form a 3:5 ratio.  So, x=(-3+9)=6.

From point a to point b, y goes from -14 to 2.  That's 16 units.  The values 6:10 form a 3:5 ratio.  So, y=(-14+6)=-8.

 

Point p is (6,-8)