Question 780900
<pre>
In a message dated 9/5/2013 9:00:27 A.M. Eastern Daylight Time, AnlytcPhil@aol.com writes:
 (0,-3), (-1/2,7/2) and (-7/2,1/2)

Plot the points, and connect them:

{{{drawing(400,400,-10,10,-10,10,graph(400,400,-10,10,-10,10),

circle(0,-3,.1), circle(-1/2,7/2,.1), circle(-7/2,1/2,.1),
circle(0,-3,.05), circle(-1/2,7/2,.05), circle(-7/2,1/2,.05),

green(line(-1/2,7/2,-7/2,1/2),line(-1/2,7/2,0,-3),line(0,-3,-7/2,1/2))
 )}}} 

These green lines are the mid-segments of the triangle we are trying
to determine the coordinates of.

We remember that a mid-segment of a triangle (which joins the 
midpoints of two sides of a triangle) is parallel to the third 
side of the triangle.  The mid-segment is also 1/2 of the third
side , but we do not need that fact.

Through each of those points we will find the equation of the line 
paralell to the mid-segment joining the other two points:

We find the slope of the mid-segment joining

(0,-3) and  ({{{-1/2}}},{{{7/2}}})

m = {{{(y[2]-y[1])/(x[2]-x[1])}}}

where (x<sub>1</sub>,y<sub>1</sub>) = (0,-3)
and where (x<sub>2</sub>,y<sub>2</sub>) = ({{{-7/2}}},{{{1/2}}})

m = {{{((7/2)-(-3))/((-1/2)-(0))}}} = {{{(7/2+3)/(-1/2-0)}}} = {{{(13/2)/(-1/2)}}} = {{{13/2}}}÷{{{-1/2}}} = {{{13/2}}}×{{{-2/1}}} = {{{13/cross(2)}}}×{{{-cross(2)/1}}} = -13

Now we find the equation of the line through the third point ({{{-7/2}}},{{{1/2}}}):

Point-slope formula:

y - y<sub>1</sub> = m(x - x<sub>1</sub>)

where (x<sub>1</sub>,y<sub>1</sub>) = the third point ({{{-7/2}}},{{{1/2}}})

y - {{{1/2}}} = -13(x - ({{{-7/2}}}))

y - {{{1/2}}} = -13(x + {{{7/2}}})

y - {{{1/2}}} = -13x - {{{91/2}}})

    y = -13x - {{{90/2}}}

    y = -13x - 45  

That line is the red one below:

{{{drawing(400,400,-10,10,-10,10,graph(400,400,-10,10,-10,10),

circle(0,-3,.1), circle(-1/2,7/2,.1), circle(-7/2,1/2,.1),
circle(0,-3,.05), circle(-1/2,7/2,.05), circle(-7/2,1/2,.05),
red(line(12,-201,-13,124)),
green(line(-1/2,7/2,-7/2,1/2),line(-1/2,7/2,0,-3),line(0,-3,-7/2,1/2))
 )}}}

-------------------

We find the slope of the mid-segment joining

(0,-3) and ({{{-7/2}}},{{{1/2}}})

m = {{{(y[2]-y[1])/(x[2]-x[1])}}}

where (x<sub>1</sub>,y<sub>1</sub>) = (0,-3)
and where (x<sub>2</sub>,y<sub>2</sub>) = ({{{-7/2}}},{{{1/2}}})

m = {{{((1/2)-(-3))/((-7/2)-(0))}}} = {{{(1/2+3)/(-7/2-0)}}} = {{{(1/2+6/2)/(-7/2)}}} = {{{(7/2)/(-7/2)}}} = {{{7/2}}}÷{{{-7/2}}} = -1

Now we find the equation of the line through the third point ({{{-1/2}}},{{{7/2}}}):

Point-slope formula:

y - y<sub>1</sub> = m(x - x<sub>1</sub>)

where (x<sub>1</sub>,y<sub>1</sub>) = the third point ({{{-1/2}}},{{{7/2}}})

y - {{{7/2}}} = -1(x - ({{{-1/2}}}))

y - {{{7/2}}} = -1(x + {{{1/2}}})

y - {{{7/2}}} = -x - {{{1/2}}})

    y = -x + {{{6/2}}}

    y = -x + 3  

 

That line is the second red one below:

{{{drawing(400,400,-10,10,-10,10,graph(400,400,-10,10,-10,10),

circle(0,-3,.1), circle(-1/2,7/2,.1), circle(-7/2,1/2,.1),
circle(0,-3,.05), circle(-1/2,7/2,.05), circle(-7/2,1/2,.05),
red(line(12,-9,-13,16)),red(line(12,-201,-13,124)),
green(line(-1/2,7/2,-7/2,1/2),line(-1/2,7/2,0,-3),line(0,-3,-7/2,1/2))
 )}}}

-------------------

We find the slope of the mid-segment joining

({{{-1/2}}},{{{7/2}}}) and ({{{-7/2}}},{{{1/2}}})

m = {{{(y[2]-y[1])/(x[2]-x[1])}}}

where (x<sub>1</sub>,y<sub>1</sub>) = (0,-3)
and where (x<sub>2</sub>,y<sub>2</sub>) = ({{{-7/2}}},{{{1/2}}})

m = {{{((1/2)-(7/2))/((-7/2)-(-1/2))}}} = {{{(1/2-7/2)/(-7/2+1/2)}}} = {{{(-6/2)/(-6/2)}}} = {{{(-3)/(-3)}}} = 1

Now we find the equation of the line through the third point  (0,-3), :

Point-slope formula:

y - y<sub>1</sub> = m(x - x<sub>1</sub>)

where (x<sub>1</sub>,y<sub>1</sub>) = the third point ({{{-1/2}}},{{{7/2}}})

y - (-3)} = 1(x - 0)

y + 3 = x 

    y = x - 3  

 

That line is the second red one below:

{{{drawing(400,400,-10,10,-10,10,graph(400,400,-10,10,-10,10),

circle(0,-3,.1), circle(-1/2,7/2,.1), circle(-7/2,1/2,.1),
circle(0,-3,.05), circle(-1/2,7/2,.05), circle(-7/2,1/2,.05),
red(line(12,-9,-13,16)),red(line(12,-201,-13,124),line(-14,-17,14,11)),
green(line(-1/2,7/2,-7/2,1/2),line(-1/2,7/2,0,-3),line(0,-3,-7/2,1/2))
 )}}}

Now we find the three points of intersection of the three pairs of
the lines

y = -13x - 45,  y = -x + 3, y = x - 3

Solve these three systems:

y = -13x - 45      y = -13x - 45      y = -x + 3   
y = -x + 3         y = x - 3          y = x - 3

You can do that.  They are

(-4,7),  (-3,-6), and (3,0)

Edwin</pre>