Question 896575
<pre>
We let the x-component of Q be a, Then since Q is on the line
{{{y=expr(1/2)x-2}}}, the y-component of Q is found by substituting
a for x in {{{y=expr(1/2)x-2}}} and get {{{y=expr(1/2)a-2}}}. So
Q is the point Q(a,{{{expr(1/2)a-2}}}),

We let the x-component of R be b, Then since R is on the line
{{{y=-x+5}}}, the y-component of R is found by substituting
b for x in {{{y=-x+5}}} and get {{{y=-b+5}}}. So
R is the point R(b,-b+5)

{{{drawing(400,400,-4,8,-4,8,graph(400,400,-4,8,-4,8),
locate(1.2,1,"P(1,1)"), line(-10,-7,12,4),line(12,-7,-11,16),

line(1.5,3.5,4/9,-16/9), locate(1.8,3.7,"R(b,-b+5)"),locate(4/9,-16/9+.1,Q(a,expr(1/2)a-2)),

circle(1.5,3.5,0.15),circle(1.5,3.5,0.13),circle(1.5,3.5,0.11),circle(1.5,3.5,0.09),circle(1.5,3.5,0.07),circle(1.5,3.5,0.05),circle(1.5,3.5,0.03),circle(1.5,3.5,0.01),locate(6,-3,y=-x+5),

locate(6,2.7,y=expr(1/2)x-2),


circle(0.44444444,-1.77777778,0.15),circle(0.44444444,-1.77777778,0.13),circle(0.44444444,-1.77777778,0.11),circle(0.44444444,-1.77777778,0.09),circle(0.44444444,-1.77777778,0.07),circle(0.44444444,-1.77777778,0.05),circle(0.44444444,-1.77777778,0.03),circle(0.44444444,-1.77777778,0.01),



circle(1,1,0.15),circle(1,1,0.13),circle(1,1,0.11),circle(1,1,0.09),circle(1,1,0.07),circle(1,1,0.05),circle(1,1,0.03),circle(1,1,0.01)  )}}}

We are told that P(1,1) is the midpoint of QR.  We use the
midpoint formula:

Midpoint = {{{(matrix(1,3,(x[1]+x[2])/2, ",",(y[1]+y[2])/2))}}}

Midpoint = {{{(matrix(1,3,(a+b)/2, ",",((expr(1/2)a-2)+(-b+5))/2))}}}

So we equate those coordinates to the coordinates of P(1,1).

Equating the x-coordinates of P

{{{(a+b)/2=1}}}
{{{a+b=2}}}

Equating the y-coordinates of P

{{{((expr(1/2)a-2)+(-b+5))/2=1}}}
{{{(expr(1/2)a-2)+(-b+5)=2}}}
{{{expr(1/2)a-2-b+5=2}}}
{{{expr(1/2)a-b+3=2}}}
Multiply through by 2:
{{{a-2b+6=4}}}
{{{a-2b=-2}}}

So we solve the system by substitution or elimination:

{{{system(a+b=2,a-2b=-2)}}}
Solve the first for a=2-b
Substitute in the second equation:
(2-b)-2b=-2
  2-b-2b=-2
    2-3b=-2
     -3b=-4
       b={{{4/3}}}

a=2-b
a=2-{{{4/3}}}
a={{{6/3-4/3}}}
a={{{2/3}}}

So the point {{{Q(a,expr(1/2)a-2))}}} becomes:

{{{Q(2/3,expr(1/2)(2/3)-2))}}}

{{{Q(2/3,1/3-2))}}}

{{{Q(2/3,1/3-6/3))}}}

{{{Q(2/3,-5/3))}}}

So the point R(b,-b+5) becomes:

{{{R(4/3,-4/3+5)}}}

{{{R(4/3,-4/3+15/3)}}}

{{{R(4/3,11/3)}}}

Edwin</pre>