all points 3 units from the x axis are on y=3 and y=-3
the distance formula is [tex]D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}[/tex]
ok, one point is (5,4) and the other is (x,3) or (x,-3), find both
for (x,3) (the ones on y=3)
[tex]3=\sqrt{(5-x)^2+(4-3)^2}[/tex]
[tex]3=\sqrt{x^2-10x+25+(1)^2}[/tex]
[tex]3=\sqrt{x^2-10x+26}[/tex]
squaer both sides
9=x²-10x+26
minus 9
0=x²-10x+17
factor
we can't
using quadratic formula
x=5-2√2 or 5+2√2
for (x,-3)
[tex]3=\sqrt{(5-x)^2+(4-(-3))^2}[/tex]
[tex]3=\sqrtx{x^2-10x+25+(7)^2}[/tex]
[tex]3=\sqrtx{x^2-10x+25+49}[/tex]
[tex]3=\sqrtx{x^2-10x+74}[/tex]
squaer both sides
9=x²-10x+74
0=x²-10x+65
solving using quadratic formula we get a false statment
no points
so the 2 points are
[tex](5+2\sqrt{2},3)[/tex] and [tex](5-2\sqrt{2},3)[/tex]
