Question 1058558
<b><font size = 5>Darn, I misread ordinate and abscissa!</b></font>

The vertices of the base of an isosceles *triangle* are 
(1,2)and R (4,-1).  Find the ordinate of the third vertex 
if its abscissa is 6.
<font size=4><pre><b>
You have this:

{{{drawing(400,4000/13,-2.5,10.5,-2.5,7.5,

line(1,2,4,-1),green(line(1,2,8,6),line(4,-1,8,6)),
locate(-.5,2.4,"(1,2)"), locate(4,-1,"(4,-1)"), locate(8.1,6.4,"(x,6)"),
line(-5,0,15,0), line(0,-5,0,15) )}}}

and you want to find x.

Since the triangle is isosceles, we use the distance formula
to set the lengths of the two green sides equal to each other:

{{{sqrt((x-1^"")^2+(6-2^"")^2)}}}{{{""=""}}}{{{sqrt((x-4^"")^2+(6-(-1)^"")^2)}}}

Square both sides. (Take away the square root symbols)

{{{(x-1^"")^2+(6-2^"")^2}}}{{{""=""}}}{{{(x-4^"")^2+(6-(-1)^"")^2}}}

Simplify:

{{{(x-1^"")^2+(4^"")^2}}}{{{""=""}}}{{{(x-4^"")^2+(6+1^"")^2}}}

{{{(x-1^"")^2+16}}}{{{""=""}}}{{{(x-4^"")^2+(7^"")^2}}}

{{{(x-1^"")^2+16}}}{{{""=""}}}{{{(x-4^"")^2+49}}}

Get the squares on the left and the number on the right:

{{{(x-1^"")^2-(x-4^"")^2}}}{{{""=""}}}{{{33}}}

Factor the left as a difference of squares

{{{((x-1)^""-(x-4)^"")((x-1)^""+(x-4)^"")}}}{{{""=""}}}{{{33}}}

{{{(x-1-x+4)(x-1+x-4)}}}{{{""=""}}}{{{33}}}

{{{3(2x-5)}}}{{{""=""}}}{{{33}}}

{{{6x-15}}}{{{""=""}}}{{{33}}}

{{{6x}}}{{{""=""}}}{{{48}}}

{{{x}}}{{{""=""}}}{{{8}}}

So the abscissa (x-coordinate) of the vertex of the triangle
is 8, and so the vertex is the point (8,6).

{{{drawing(400,4000/13,-2.5,10.5,-2.5,7.5,

line(1,2,4,-1),green(line(1,2,8,6),line(4,-1,8,6)),
locate(-.5,2.4,"(1,2)"), locate(4,-1,"(4,-1)"), locate(8.1,6.4,"(8,6)"),
line(-5,0,15,0), line(0,-5,0,15) )}}}

Edwin</pre></b>