Question 941945
<pre>
Find the point on the x-axis that is equidistant from (-4,-3) and (-1,5)

{{{drawing(300,400,-6,3,-5,7, line(-10,0,10,0), line(0,-10,0,10),
green(line(1/6,0,-1,5),line(1/6,0,-4,-3)), locate(1/6,0,"(x,0)"),


circle(-1,5,0.15),circle(-1,5,0.13),circle(-1,5,0.11),circle(-1,5,0.09),circle(-1,5,0.07),circle(-1,5,0.05),circle(-1,5,0.03),circle(-1,5,0.01),

circle(-4,-3,0.15),circle(-4,-3,0.13),circle(-4,-3,0.11),circle(-4,-3,0.09),circle(-4,-3,0.07),circle(-4,-3,0.05),circle(-4,-3,0.03),circle(-4,-3,0.01),

circle(0.16666667,0,0.09),circle(0.16666667,0,0.07),circle(0.16666667,0,0.05),circle(0.16666667,0,0.03),circle(0.16666667,0,0.01),

locate(-2,5.7,"(-1,5)"), locate(-4,-3,"(-4,-3)")



 )}}}

Every point on the x-axis has its y-coordinate as 0.  So we label
the point we are looking for (x,0).

Then we use the distance formula to set the distances from (x,0) to
the two given points equal to each other:

{{{sqrt((x^""-(-1))^2+(0-5)^2)}}}{{{""=""}}}{{{sqrt((x^""-(-4))^2+(0-(-3))^2)}}}

{{{sqrt((x+1)^2+(-5)^2)}}}{{{""=""}}}{{{sqrt((x+4)^2+(0+3)^2)}}}

{{{sqrt((x+1)^2+25)}}}{{{""=""}}}{{{sqrt((x+4)^2+(3)^2)}}}

{{{sqrt((x+1)^2+25)}}}{{{""=""}}}{{{sqrt((x+4)^2+9)}}}

{{{(x+1)^2+25}}}{{{""=""}}}{{{(x+4)^2+9}}}

{{{x^2+2x+1+25}}}{{{""=""}}}{{{x^2+8x+16+9}}}

{{{x^2+2x+26}}}{{{""=""}}}{{{x^2+8x+25}}}

{{{2x+26}}}{{{""=""}}}{{{8x+25}}}

{{{1}}}{{{""=""}}}{{{6x}}}

{{{1/6}}}{{{""=""}}}{{{x}}}

So the point on the x-axis is {{{(matrix(1,3,1/6,",",0))}}}

Edwin</pre>