Question 1020439
<pre><b>

{{{drawing(400,400,-3.5,10.5,-6,8,

locate(5,6.7,"A(5,6)"), locate(9.2,3,C), locate(.26,-3.8,"B(0,-4)"),
locate(4.7,-.5,"D(h,k)"),
graph(400,400,-3.5,10.5,-6,8),

line(0,-4,5,6), line(0,-4,8.94427191,2.708203932),
line(5,6,8.94427191,2.708203932),


circle(5,6,0.15),circle(5,6,0.13),circle(5,6,0.11),circle(5,6,0.09),circle(5,6,0.07),circle(5,6,0.05),circle(5,6,0.03),circle(5,6,0.01),

circle(0,-4,0.15),circle(0,-4,0.13),circle(0,-4,0.11),circle(0,-4,0.09),circle(0,-4,0.07),circle(0,-4,0.05),circle(0,-4,0.03),circle(0,-4,0.01),

circle(4.47213596,-0.64589803,0.15),circle(4.47213596,-0.64589803,0.13),circle(4.47213596,-0.64589803,0.11),circle(4.47213596,-0.64589803,0.09),circle(4.47213596,-0.64589803,0.07),circle(4.47213596,-0.64589803,0.05),circle(4.47213596,-0.64589803,0.03),circle(4.47213596,-0.64589803,0.01),

circle(8.94427191,2.70820393,0.15),circle(8.94427191,2.70820393,0.13),circle(8.94427191,2.70820393,0.11),circle(8.94427191,2.70820393,0.09),circle(8.94427191,2.70820393,0.07),circle(8.94427191,2.70820393,0.05),circle(8.94427191,2.70820393,0.03),circle(8.94427191,2.70820393,0.01) )}}}

Let the unknown point D be D(h,k).

We can find the equation of AD because we are given
its gradient 3/4 and its y-intercept (0,-4)

{{{matrix(1,5,

y,""="",(matrix(3,1,gradient,of,line))x,""+"",(matrix(3,1,y-coordinate,of,y-intercept)))}}}

{{{matrix(1,5,y,""="",expr(3/4)x,""+"",(-4))}}}

{{{matrix(1,5,y,""="",expr(3/4)x,""-"",4)}}}

That's the equation of AD.

Next we find the length of AB using the distance formula:


{{{d=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}

{{{AB=sqrt(((6)-(-4)^"")^2+((5)-(0))^2)}}}

{{{AB=sqrt((6+4)^2+(5-0)^2)}}}

{{{AB=sqrt((10)^2+(5)^2)}}}

{{{AB=sqrt(100+25)}}}

{{{AB=sqrt(125)}}}

{{{AB=sqrt(25*5)}}}

{{{AB=5sqrt(5)}}}

Since triangle ABC is isosceles, we know that

{{{BC=AB=5sqrt(5)}}}

And since D is the midpoint of BC, then AD is
one-half of BC, so

{{{BD=5sqrt(5)/2}}}

We use the distance formula and set the length 
of BD equal to this:

{{{d=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}

Square both sides of the equation:

{{{d^2=(x[2]-x[1])^2+(y[2]-y[1])^2}}}

So we have this equation, setting {{{5sqrt(5)/2}}} 
equal to the distance from B(0,-4) to D(h,k):

{{{(5sqrt(5)/2)^2=((h)-(0)^"")^2+((k)^""-(-4))^2}}}

{{{(25*5)/4=(h)^2+(k+4)^2}}}

{{{125/4=h^2+(k+4)^2}}}

Multiply through by 4 to clear the fraction:

{{{125=4h^2+4(k+4)^2}}}

Since we know that D(h,k) is a point on line AD,
whose equation is 

{{{matrix(1,5,y,""="",expr(3/4)x,""-"",4)}}}

So D(h,k) it must satisfy that equation: 

{{{k=expr(3/4)h-4}}}.

So we substitute that for k in

{{{125=4h^2+4(k+4)^2}}}

{{{125=4h^2+4(expr(3/4)h-4+4)^2}}}

{{{125=4h^2+4(expr(3/4)h)^2}}}

{{{125=4h^2+4*expr(9/16)h^2}}}

{{{125=4h^2+expr(9/4)h^2}}}

Multiply through by 4

{{{500=16h^2+9h^2)}}}

{{{500=25h^2}}}

{{{20=h^2}}}

{{{sqrt(20)=h}}}

{{{sqrt(4*5)=h}}}

{{{2sqrt(5)=h}}}

{{{k=expr(3/4)h-4}}}

{{{k=expr(3/4)(2sqrt(5))-4}}}

{{{k=expr(3/4)(2sqrt(5))-4}}}

{{{k=expr(3/2)sqrt(5)-4}}}

{{{k=(3sqrt(5)-8)/2}}}

So the coordinates of point D is

{{{(matrix(1,3,2sqrt(5),",",(3sqrt(5)-8)/2))}}}

Edwin</pre></b>