Question 870669
{{{drawing(300,300,-2,10,-2,10,circle(7,2,0.2),circle(1,4,0.2),line(7,2,1,4))}}}
Find the slope of the line connecting A and C.
{{{m=(4-2)/(1-7)=2/-6=-1/3}}}
Diagonal BD is perpendicular to diagonal AC so their slopes are negative reciprocals of each other.
{{{m[perp]*(-1/3)=-1}}}
{{{m{perp]=3}}}
Diagonal BD also goes through the midpoint of AC. 
{{{x[m]=(7+1)/2=8/2=4}}}
{{{y[m]=(2+4)/2=6/2=3}}}
Use this point to find the line, BC, {{{y=3x+b}}}
{{{3=3(4)+b}}}
{{{b=-9}}}
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{{{y=3x-9}}}
{{{drawing(300,300,-2,10,-2,10,circle(7,2,0.2),circle(1,4,0.2),line(7,2,1,4),graph(300,300,-2,10,-2,10,3x-9))}}}
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The distance from the midpoint to points B and D is the same as the distance from the midpoint to A and C.
{{{D^2=(7-4)^2+(2-3)^2}}}
{{{D^2=9+1}}}
{{{D^2=10}}}
The distance from the midpoint to B and D is,
{{{10=(x-4)^2+(y-3)^2}}}
But you also know that B and D lie on the line {{{y=3x-9}}}
So substituting.
{{{10=(x-4)^2+(3x-9-3)^2}}}
{{{10=x^2-8x+16+9x^2-72x+144}}}
{{{10=10x^2-80x+160}}}
{{{10x^2-80x+150=0}}}
{{{x^2-8x+15=0}}}
{{{(x-3)(x-5)=0}}}
Two solutions:
{{{x-3=0}}}
{{{x=3}}}
Then
{{{y=3(3)-9}}}
{{{y=0}}}
and
{{{x-5=0}}}
{{{x=5}}}
Then,
{{{y=3(5)-9}}}
{{{y=6}}}
B and D are (3,0) and (5,6)
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{{{drawing(300,300,-2,10,-2,10,circle(7,2,0.2),circle(1,4,0.2),circle(3,0,0.2),circle(5,6,0.2),line(1,4,3,0),line(3,0,7,2),line(7,2,5,6),line(5,6,1,4),graph(300,300,-2,10,-2,10,3x-9))}}}