Question 152310
26.) The midpoint of the line segment joining a moving point to (6,0) is on the line y=x. Find the equation of its locus.
Ans. x-y+6=0

<pre><font size = 4 color="indigo"><b>

Graph the line {{{y=x}}}:

{{{graph(400,400,-7,13,-7,13,x)}}}

Call point (6,0) the point R.

Draw an arbitrary line from point R(6,0), crossing the
line {{{y=x}}} at point M, and extending to P so that MP
equals RM. making M the midpoint of RP.  P is the moving
point.

{{{drawing(400,400,-7,13,-7,13, locate(4.9,4.7,"M"),locate(6,1,"R(6,0)"),
graph(400,400,-7,13,-7,13,x),line(6,0,3,9), locate(2,9.7,"P")  )}}}

Label the moving point P with the variable coordinates (x,y) 

{{{drawing(400,400,-7,13,-7,13, locate(4.9,4.7,"M"),locate(6,1,"R(6,0)"),
graph(400,400,-7,13,-7,13,x),line(6,0,3,9), locate(2,9.7,"P(x,y)")  )}}}

Since point M is the midpoint of the line segment RP, we use
the midpoint formula to label M with the coordinates 
({{{(x+6)/2}}},{{{(y+0)/2}}}) or ({{{(x+6)/2}}},{{{y/2}}}) 

{{{drawing(400,400,-7,13,-7,13, locate(4.9,4.7,"M((x+6)/2,y/2)"),locate(6,1,"R(6,0)"),
graph(400,400,-7,13,-7,13,x),line(6,0,3,9), locate(2,9.7,"P(x,y)")  )}}}

Now the point M({{{(x+6)/2}}},{{{y/2}}}) is on the line {{{y=x}}}, so
it must satisfy the equation {{{y=x}}}, so we substitute that point 
into the equation {{{y=x}}} and we get:

{{{y/2=(x+6)/2}}}

Multiply both sides by 2:

{{{y = x+6}}}

That is equivalent to

{{{x-y+6=0}}}

To see the path of the moving point we draw the graph of
{{{x-y+6=0}}} in green

{{{drawing(400,400,-7,13,-7,13, locate(4.9,4.7,"M((x+6)/2,y/2)"),locate(6,1,"R(6,0)"),
graph(400,400,-7,13,-7,13,x,x+6),line(6,0,3,9), locate(2,9.7,"P(x,y)")  )}}}


Edwin</pre>