Question 926264
Given the points A(-2, 1) and B(4, 5), determine the coordinates of point P on directed line segment that partitions in the ratio 7/3.
<pre>

{{{ratio}}}{{{""=""}}}{{{
matrix(1,4,LONG,PART,OF,AB)
/ 
matrix(1,4,SHORT,PART,OF,AB)

=7/3 }}}

The ratio of the parts is not k.  We calculate k this way:

{{{k}}}{{{""=""}}}{{{(matrix(9,1,the, fraction, which, the, long, part, is, of, AB))}}}{{{""=""}}}{{{matrix(1,3,

matrix(1,4,LONG,PART,OF,AB)/matrix(1,9,LONG,PART,OF,AB,""+"",SHORT,PART,OF,AB),
""="",
7/(7+3)=7/10)
)}}}

That is, to find k we must add the numerator to the denominator of the
ratio of the two parts.  We plot the point and draw the line connecting them,
guessing about where P(?,?) might be:

{{{drawing(400,360,-4,6,-2,7,grid(1), graph(400,360,-4,6,-2,7),line(4,5,-2,1), locate(-3,1,"A(-2,1)"), locate(4.3,5,"B(4, 5)"),
circle(2.2,3.8,0.15),circle(2.2,3.8,0.13),circle(2.2,3.8,0.11),circle(2.2,3.8,0.09),circle(2.2,3.8,0.07),circle(2.2,3.8,0.05),circle(2.2,3.8,0.03),circle(2.2,3.8,0.01),locate(2.5,3.8,"P(?,?)") 



)}}}

Now we draw the run (in green) and the rise (in red):

{{{drawing(400,360,-4,6,-2,7,grid(1), graph(400,360,-4,6,-2,7),line(4,5,-2,1), 
green(line(-2,1,4,1)), red(line(4,1,4,5)),green(locate(1.2,1,RUN)),
red(locate(4.1,3,RISE)),

circle(2.2,3.8,0.15),circle(2.2,3.8,0.13),circle(2.2,3.8,0.11),circle(2.2,3.8,0.09),circle(2.2,3.8,0.07),circle(2.2,3.8,0.05),circle(2.2,3.8,0.03),circle(2.2,3.8,0.01),locate(2.5,3.8,"P(?,?)"), 

locate(-3,1,"A(-2,1)"), locate(4.3,5,"B(4, 5)") )}}}

Now we count the blocks and find that the green RUN is 6 units long
and that the red RISE is 4 units long.

{{{drawing(400,360,-4,6,-2,7,grid(1), graph(400,360,-4,6,-2,7),line(4,5,-2,1), 
green(line(-2,1,4,1)), red(line(4,1,4,5)),green(locate(1.2,1,RUN=6)),
red(locate(4.1,3,RISE=4)),circle(2.2,3.8,0.15),circle(2.2,3.8,0.13),circle(2.2,3.8,0.11),circle(2.2,3.8,0.09),circle(2.2,3.8,0.07),circle(2.2,3.8,0.05),circle(2.2,3.8,0.03),circle(2.2,3.8,0.01),locate(2.5,3.8,"P(?,?)"), 

locate(-3,1,"A(-2,1)"), locate(4.3,5,"B(4, 5)") )}}}

Now since we know the RUN=6 and the RISE=4, and that 

k = 7/10, x1=[x-coordinate of A]=-2, y1=[y-coordinate of A]=1, we are
ready to substitute:

The x-coordinate of P is x1 + k • run = -2 + 7/10 • 6 = 

      -2 + 42/10 = -2 + 21/5 = -10/5 + 21/5 = 11/5        

The y-coordinate of P is y1 + k • rise = 1 + 7/10 • 4 = 

      1 + 28/10 = 1 + 14/5 = 5/5 + 14/5 = 19/5   


The coordinates of P are (11/5,19/5).  Final graph:

{{{drawing(400,360,-4,6,-2,7,grid(1), graph(400,360,-4,6,-2,7),line(4,5,-2,1), 
green(line(-2,1,4,1)), red(line(4,1,4,5)),green(locate(1.2,1,RUN=6)),
red(locate(4.1,3,RISE=4)),circle(2.2,3.8,0.15),circle(2.2,3.8,0.13),circle(2.2,3.8,0.11),circle(2.2,3.8,0.09),circle(2.2,3.8,0.07),circle(2.2,3.8,0.05),circle(2.2,3.8,0.03),circle(2.2,3.8,0.01),locate(2.5,3.8,P(11/5,19/5)), 

locate(-3,1,"A(-2,1)"), locate(4.3,5,"B(4, 5)") )}}}

Edwin</pre>